help with sum of infinite series, stuck in problem The Next CEO of Stack OverflowComparing series with convergent seriesHow to calculate the sum of the infinite series $sum_n = 0^infty fracn2^sqrtn$Calculating the sum of an infinite series (high school calc)Infinite sum of alternating telescoping seriesHelp understanding the sequence of partial sums of a seriesSum of an infinite series $(1 - frac 12) + (frac 12 - frac 13) + cdots$ - not geometric series?Definition of series and their relation to sequences and functionsLimit of an infinite series as limit of sequence of partial sums.Convergence test for partial sum whose elements all change as the index increasesLimit Of Infinite Series
If the heap is initialized for security, then why is the stack uninitialized?
What's the point of interval inversion?
Anatomically Correct Strange Women In Ponds Distributing Swords
How can I open an app using Terminal?
How do I solve this limit?
Why do remote companies require working in the US?
How should I support this large drywall patch?
Why here is plural "We went to the movies last night."
How do I construct this japanese bowl?
How to write the block matrix in LaTex?
What happens if you roll doubles 3 times then land on "Go to jail?"
Is it safe to use c_str() on a temporary string?
Is a stroke of luck acceptable after a series of unfavorable events?
How to count occurrences of text in a file?
Whats the best way to handle refactoring a big file?
Go Pregnant or Go Home
Why do professional authors make "consistency" mistakes? And how to avoid them?
How to be diplomatic in refusing to write code that breaches the privacy of our users
What is the point of a new vote on May's deal when the indicative votes suggest she will not win?
How do I go from 300 unfinished/half written blog posts, to published posts?
Fastest way to shutdown Ubuntu Mate 18.10
Would this house-rule that treats advantage as a +1 to the roll instead (and disadvantage as -1) and allows them to stack be balanced?
Opposite of a diet
When airplanes disconnect from a tanker during air to air refueling, why do they bank so sharply to the right?
help with sum of infinite series, stuck in problem
The Next CEO of Stack OverflowComparing series with convergent seriesHow to calculate the sum of the infinite series $sum_n = 0^infty fracn2^sqrtn$Calculating the sum of an infinite series (high school calc)Infinite sum of alternating telescoping seriesHelp understanding the sequence of partial sums of a seriesSum of an infinite series $(1 - frac 12) + (frac 12 - frac 13) + cdots$ - not geometric series?Definition of series and their relation to sequences and functionsLimit of an infinite series as limit of sequence of partial sums.Convergence test for partial sum whose elements all change as the index increasesLimit Of Infinite Series
$begingroup$
the series in the problem is as follows, and we would like to see if the series converges, and what is the sum of infinite series:
$$sum_n=1^infty frac2^n+1+13^n$$
It's a homework problem, but I already asked my teacher about it at school and he didn't really help me that much.
The earlier problems that we had were easier than this one, because I cannot find the common ratio for this infinite series (infinite geometric series should have common ratio, and when you know it you can tell if there is convergence)
I read online after class, that what I should do is to find the limit of the sequence of partial sums.
This limit, if it exists and is a real-number, should be the value of the sum of the infinite series.
It makes sense conceptually when you have a large value of k, for a finite series $S_k$, when k becomes larger the finite sum becomes more and more like the sum of infinite series.
So that if I have partial sums in a sequence $S_1, S_2, S_3...S_k $
The larger the k, the better it is representing infinite series, assuming it converges at all, towards anything.
I managed to put some large values of n, into wolfram alpha and the correct answer seems to be that the infinite series converges and sum of infinite series is 4.5
But I don't have a solid idea how I could find the explicit formula for the sequence of partial sums for this particular infinite series. And I don't particularly have any other good ideas how to prove that the sum of infinite series should be 4.5 in this case...
sequences-and-series convergence summation
edited yesterday
José Carlos Santos
171k23132240
asked Jan 25 '18 at 15:40
Late347Late347
366
$endgroup$
add a comment |
$begingroup$
the series in the problem is as follows, and we would like to see if the series converges, and what is the sum of infinite series:
$$sum_n=1^infty frac2^n+1+13^n$$
It's a homework problem, but I already asked my teacher about it at school and he didn't really help me that much.
The earlier problems that we had were easier than this one, because I cannot find the common ratio for this infinite series (infinite geometric series should have common ratio, and when you know it you can tell if there is convergence)
I read online after class, that what I should do is to find the limit of the sequence of partial sums.
This limit, if it exists and is a real-number, should be the value of the sum of the infinite series.
It makes sense conceptually when you have a large value of k, for a finite series $S_k$, when k becomes larger the finite sum becomes more and more like the sum of infinite series.
So that if I have partial sums in a sequence $S_1, S_2, S_3...S_k $
The larger the k, the better it is representing infinite series, assuming it converges at all, towards anything.
I managed to put some large values of n, into wolfram alpha and the correct answer seems to be that the infinite series converges and sum of infinite series is 4.5
But I don't have a solid idea how I could find the explicit formula for the sequence of partial sums for this particular infinite series. And I don't particularly have any other good ideas how to prove that the sum of infinite series should be 4.5 in this case...
sequences-and-series convergence summation
edited yesterday
José Carlos Santos
171k23132240
asked Jan 25 '18 at 15:40
Late347Late347
366
$endgroup$
add a comment |
$begingroup$
the series in the problem is as follows, and we would like to see if the series converges, and what is the sum of infinite series:
$$sum_n=1^infty frac2^n+1+13^n$$
It's a homework problem, but I already asked my teacher about it at school and he didn't really help me that much.
The earlier problems that we had were easier than this one, because I cannot find the common ratio for this infinite series (infinite geometric series should have common ratio, and when you know it you can tell if there is convergence)
I read online after class, that what I should do is to find the limit of the sequence of partial sums.
This limit, if it exists and is a real-number, should be the value of the sum of the infinite series.
It makes sense conceptually when you have a large value of k, for a finite series $S_k$, when k becomes larger the finite sum becomes more and more like the sum of infinite series.
So that if I have partial sums in a sequence $S_1, S_2, S_3...S_k $
The larger the k, the better it is representing infinite series, assuming it converges at all, towards anything.
I managed to put some large values of n, into wolfram alpha and the correct answer seems to be that the infinite series converges and sum of infinite series is 4.5
But I don't have a solid idea how I could find the explicit formula for the sequence of partial sums for this particular infinite series. And I don't particularly have any other good ideas how to prove that the sum of infinite series should be 4.5 in this case...
sequences-and-series convergence summation
edited yesterday
José Carlos Santos
171k23132240
asked Jan 25 '18 at 15:40
Late347Late347
366
$endgroup$
the series in the problem is as follows, and we would like to see if the series converges, and what is the sum of infinite series:
$$sum_n=1^infty frac2^n+1+13^n$$
It's a homework problem, but I already asked my teacher about it at school and he didn't really help me that much.
The earlier problems that we had were easier than this one, because I cannot find the common ratio for this infinite series (infinite geometric series should have common ratio, and when you know it you can tell if there is convergence)
I read online after class, that what I should do is to find the limit of the sequence of partial sums.
This limit, if it exists and is a real-number, should be the value of the sum of the infinite series.
It makes sense conceptually when you have a large value of k, for a finite series $S_k$, when k becomes larger the finite sum becomes more and more like the sum of infinite series.
So that if I have partial sums in a sequence $S_1, S_2, S_3...S_k $
The larger the k, the better it is representing infinite series, assuming it converges at all, towards anything.
I managed to put some large values of n, into wolfram alpha and the correct answer seems to be that the infinite series converges and sum of infinite series is 4.5
But I don't have a solid idea how I could find the explicit formula for the sequence of partial sums for this particular infinite series. And I don't particularly have any other good ideas how to prove that the sum of infinite series should be 4.5 in this case...
sequences-and-series convergence summation
sequences-and-series convergence summation
edited yesterday
José Carlos Santos
171k23132240
asked Jan 25 '18 at 15:40
Late347Late347
366
edited yesterday
José Carlos Santos
171k23132240
asked Jan 25 '18 at 15:40
Late347Late347
366
edited yesterday
José Carlos Santos
171k23132240
edited yesterday
José Carlos Santos
171k23132240
edited yesterday
José Carlos Santos
171k23132240
171k23132240
asked Jan 25 '18 at 15:40
Late347Late347
366
asked Jan 25 '18 at 15:40
Late347Late347
366
asked Jan 25 '18 at 15:40
Late347Late347
366
366
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$sum_n=1^inftyfrac2^n+1+13^n=2sum_n=1^inftyleft(frac23right)^n+sum_n=1^inftyleft(frac13right)^n.$$Can you take it from here?
answered Jan 25 '18 at 15:43
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
Can you clarify where you get the idea to split the series like that, I'm not very good at the calculation rules for sums (sigma)... Indeed I only really studied about infinite series last week. I mean does there exist some algebraic formulas for manipulation of sums like that
$endgroup$
– Late347
Jan 25 '18 at 15:55
$begingroup$
@Late347 Yes: if both series $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$ converge, then the series $sum_n=1^infty(a_n+b_n)$ converges too and its sum is the sum of the two original series.
$endgroup$
– José Carlos Santos
Jan 25 '18 at 15:57
add a comment |
$begingroup$
After 20 years, I still have to rederive the identity
$$ sum_n=1^infty alpha^n = fracalpha1-alpha tag1$$
whenever I need it. Since it isn't that difficult, perhaps we should do that first.
Suppose that $|alpha| < 1$, and let
$$ S_k := sum_n=1^k alpha^k = alpha + alpha^2 + dotsb + alpha^k $$
be the $k$-th partial sum. Then
$$ alpha S_k = sum_n=1^k alpha^k+1 = alpha^2 + alpha^3 + dotsb + alpha^k+1. $$
Subtracting, we obtain
beginalign
(1-alpha) S_k
&= S_k - alpha S_k \
&= left( alpha + alpha^2 + dotsb + alpha^k right) - left( alpha^2 + alpha^3 + dotsb + alpha^k+1 right) \
&= alpha + left( alpha^2 - alpha^2right) + left( alpha^3 - alpha^3right) + dotsb + left(alpha^k - alpha^kright) - alpha^k+1 \
&= alpha - alpha^k+1.
endalign
Solving for $S_k$, we get
$$ S_k = fracalpha - alpha^k+11-alpha. $$
Since $|alpha| < 1$, it follows that $lim_ktoinfty alpha^k+1 = 0$, and so
$$
sum_n=1^infty alpha^n
:= lim_kto infty sum_n=1^k alpha^n
= lim_ktoinfty fracalpha - alpha^k+11-alpha
= fracalpha1-alpha.
$$
But this is exactly the identity at (1). Huzzah!
Now, how do we use this here?
Observe that
$$ frac2^n+1 + 13^n
= frac2^n+13^n + frac13^n
= 2 left( frac23 right)^n + left( frac13 right)^n.
tag2$$
But series are linear, and so play nice with addition and multiplication. Specifically, if $sum a_n$ and $sum b_n$ converge, and $C$ is any constant, then we have
$$ sum_n=1^infty left[ a_n + b_n right] = sum_n=1^infty a_n + sum_n=1^infty b_n
qquadtextandqquad
sum_n=1^infty Ca_n = C sum_n=1^infty a_n. $$
These are basic rules for "simplifying" series, which are going to be quite useful for you. If you are familiar with integration, you can perform exactly these manipulations if you replace the $sum$ with $int$, and the sequences with functions. If you are not familiar with integration, ignore this comment for now, but maybe keep it in the back of your mind for later.
beginalign
sum_n=1^infty frac2^n+1 + 13^n
&= sum_n=1^infty left[ 2 left( frac23 right)^n + left( frac13 right)^n right] && (textby application of (2)) \
&= 2sum_n=1^infty left( frac23 right)^n + sum_n=1^infty left( frac13 right)^n && (textsimplify the series) \
&= 2 cdot fracfrac231-frac23 + fracfrac131- frac13 && (textby application of (1)) \
&= 2 cdot fracfrac231-frac23cdot frac33 + fracfrac131- frac13cdot frac33 && (textsilly, pedantic arithmetic) \
&= 2 cdot frac23-2 + frac13-1 && (textmore pedantic arithmetic) \
&= frac41 + frac12 && (textand some more) \
&= frac8+12 && (textalmost there) \
&= frac92. && (textdone!)
endalign
answered Jan 25 '18 at 16:02
Xander HendersonXander Henderson
14.9k103555
$endgroup$
$begingroup$
IT looks very understandable now, I jotted down my own version on a piece of paper here at my desk. My worries and further question is more about the justification of splitting the Series into two separate series... Doesn't there have to be the assumption such that the combined original series has to be convergent to begin with before manipulating? $ sum_n=1^infty frac2^n+1+13^n$
$endgroup$
– Late347
Jan 25 '18 at 16:31
$begingroup$
EDIT I just wanted to ask for sure, because I don't want to do silly mistakes with the infinite series... I still have a bad taste in my own mouth after having watched the numberphile video about summing divergent infinite series into each other (the video can be found on youtube with "1+2+3... = -1/12")
$endgroup$
– Late347
Jan 25 '18 at 16:33
$begingroup$
The justification kind of comes in through the back door. If $sum a_n$ and $sum b_n$ are convergent, then $sum (a_n + b_n)$ converges, and $sum (a_n + b_n) = sum a_n + sum b_n$. In the argument above, we break up a series of the form $sum (a_n + b_n)$ into two different convergent series $sum a_n$ and $sum b_n$, then use the fact that both of these converge in order to show that the original series converges.
$endgroup$
– Xander Henderson
Jan 25 '18 at 16:39
$begingroup$
On the other hand, if all you need to know is that the original series converges (and this might assuage your conscience regarding this assumption), you might note that $2^n+1 + 1 < 2^n+2$ for all $n$, and $sum frac2^n+23^n$ is a convergent geometric series. But then $$ 0 < sum frac2^n+1+13^n le sum frac2^n+23^n < infty, $$ which demonstrates that the original series does, in fact, converge.
$endgroup$
– Xander Henderson
Jan 25 '18 at 16:43
add a comment |
$begingroup$
You can split the sum and use
$$sum_n=1^infty r^n=fracr1-r$$
answered Jan 25 '18 at 15:46
gimusigimusi
93.1k84594
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2620771%2fhelp-with-sum-of-infinite-series-stuck-in-problem%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$sum_n=1^inftyfrac2^n+1+13^n=2sum_n=1^inftyleft(frac23right)^n+sum_n=1^inftyleft(frac13right)^n.$$Can you take it from here?
answered Jan 25 '18 at 15:43
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
Can you clarify where you get the idea to split the series like that, I'm not very good at the calculation rules for sums (sigma)... Indeed I only really studied about infinite series last week. I mean does there exist some algebraic formulas for manipulation of sums like that
$endgroup$
– Late347
Jan 25 '18 at 15:55
$begingroup$
@Late347 Yes: if both series $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$ converge, then the series $sum_n=1^infty(a_n+b_n)$ converges too and its sum is the sum of the two original series.
$endgroup$
– José Carlos Santos
Jan 25 '18 at 15:57
add a comment |
$begingroup$
$$sum_n=1^inftyfrac2^n+1+13^n=2sum_n=1^inftyleft(frac23right)^n+sum_n=1^inftyleft(frac13right)^n.$$Can you take it from here?
answered Jan 25 '18 at 15:43
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
Can you clarify where you get the idea to split the series like that, I'm not very good at the calculation rules for sums (sigma)... Indeed I only really studied about infinite series last week. I mean does there exist some algebraic formulas for manipulation of sums like that
$endgroup$
– Late347
Jan 25 '18 at 15:55
$begingroup$
@Late347 Yes: if both series $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$ converge, then the series $sum_n=1^infty(a_n+b_n)$ converges too and its sum is the sum of the two original series.
$endgroup$
– José Carlos Santos
Jan 25 '18 at 15:57
add a comment |
$begingroup$
$$sum_n=1^inftyfrac2^n+1+13^n=2sum_n=1^inftyleft(frac23right)^n+sum_n=1^inftyleft(frac13right)^n.$$Can you take it from here?
answered Jan 25 '18 at 15:43
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$$sum_n=1^inftyfrac2^n+1+13^n=2sum_n=1^inftyleft(frac23right)^n+sum_n=1^inftyleft(frac13right)^n.$$Can you take it from here?
answered Jan 25 '18 at 15:43
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 25 '18 at 15:43
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 25 '18 at 15:43
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 25 '18 at 15:43
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
$begingroup$
Can you clarify where you get the idea to split the series like that, I'm not very good at the calculation rules for sums (sigma)... Indeed I only really studied about infinite series last week. I mean does there exist some algebraic formulas for manipulation of sums like that
$endgroup$
– Late347
Jan 25 '18 at 15:55
$begingroup$
@Late347 Yes: if both series $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$ converge, then the series $sum_n=1^infty(a_n+b_n)$ converges too and its sum is the sum of the two original series.
$endgroup$
– José Carlos Santos
Jan 25 '18 at 15:57
add a comment |
$begingroup$
Can you clarify where you get the idea to split the series like that, I'm not very good at the calculation rules for sums (sigma)... Indeed I only really studied about infinite series last week. I mean does there exist some algebraic formulas for manipulation of sums like that
$endgroup$
– Late347
Jan 25 '18 at 15:55
$begingroup$
@Late347 Yes: if both series $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$ converge, then the series $sum_n=1^infty(a_n+b_n)$ converges too and its sum is the sum of the two original series.
$endgroup$
– José Carlos Santos
Jan 25 '18 at 15:57
$begingroup$
Can you clarify where you get the idea to split the series like that, I'm not very good at the calculation rules for sums (sigma)... Indeed I only really studied about infinite series last week. I mean does there exist some algebraic formulas for manipulation of sums like that
$endgroup$
– Late347
Jan 25 '18 at 15:55
$begingroup$
Can you clarify where you get the idea to split the series like that, I'm not very good at the calculation rules for sums (sigma)... Indeed I only really studied about infinite series last week. I mean does there exist some algebraic formulas for manipulation of sums like that
$endgroup$
– Late347
Jan 25 '18 at 15:55
$begingroup$
@Late347 Yes: if both series $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$ converge, then the series $sum_n=1^infty(a_n+b_n)$ converges too and its sum is the sum of the two original series.
$endgroup$
– José Carlos Santos
Jan 25 '18 at 15:57
$begingroup$
@Late347 Yes: if both series $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$ converge, then the series $sum_n=1^infty(a_n+b_n)$ converges too and its sum is the sum of the two original series.
$endgroup$
– José Carlos Santos
Jan 25 '18 at 15:57
add a comment |
$begingroup$
After 20 years, I still have to rederive the identity
$$ sum_n=1^infty alpha^n = fracalpha1-alpha tag1$$
whenever I need it. Since it isn't that difficult, perhaps we should do that first.
Suppose that $|alpha| < 1$, and let
$$ S_k := sum_n=1^k alpha^k = alpha + alpha^2 + dotsb + alpha^k $$
be the $k$-th partial sum. Then
$$ alpha S_k = sum_n=1^k alpha^k+1 = alpha^2 + alpha^3 + dotsb + alpha^k+1. $$
Subtracting, we obtain
beginalign
(1-alpha) S_k
&= S_k - alpha S_k \
&= left( alpha + alpha^2 + dotsb + alpha^k right) - left( alpha^2 + alpha^3 + dotsb + alpha^k+1 right) \
&= alpha + left( alpha^2 - alpha^2right) + left( alpha^3 - alpha^3right) + dotsb + left(alpha^k - alpha^kright) - alpha^k+1 \
&= alpha - alpha^k+1.
endalign
Solving for $S_k$, we get
$$ S_k = fracalpha - alpha^k+11-alpha. $$
Since $|alpha| < 1$, it follows that $lim_ktoinfty alpha^k+1 = 0$, and so
$$
sum_n=1^infty alpha^n
:= lim_kto infty sum_n=1^k alpha^n
= lim_ktoinfty fracalpha - alpha^k+11-alpha
= fracalpha1-alpha.
$$
But this is exactly the identity at (1). Huzzah!
Now, how do we use this here?
Observe that
$$ frac2^n+1 + 13^n
= frac2^n+13^n + frac13^n
= 2 left( frac23 right)^n + left( frac13 right)^n.
tag2$$
But series are linear, and so play nice with addition and multiplication. Specifically, if $sum a_n$ and $sum b_n$ converge, and $C$ is any constant, then we have
$$ sum_n=1^infty left[ a_n + b_n right] = sum_n=1^infty a_n + sum_n=1^infty b_n
qquadtextandqquad
sum_n=1^infty Ca_n = C sum_n=1^infty a_n. $$
These are basic rules for "simplifying" series, which are going to be quite useful for you. If you are familiar with integration, you can perform exactly these manipulations if you replace the $sum$ with $int$, and the sequences with functions. If you are not familiar with integration, ignore this comment for now, but maybe keep it in the back of your mind for later.
beginalign
sum_n=1^infty frac2^n+1 + 13^n
&= sum_n=1^infty left[ 2 left( frac23 right)^n + left( frac13 right)^n right] && (textby application of (2)) \
&= 2sum_n=1^infty left( frac23 right)^n + sum_n=1^infty left( frac13 right)^n && (textsimplify the series) \
&= 2 cdot fracfrac231-frac23 + fracfrac131- frac13 && (textby application of (1)) \
&= 2 cdot fracfrac231-frac23cdot frac33 + fracfrac131- frac13cdot frac33 && (textsilly, pedantic arithmetic) \
&= 2 cdot frac23-2 + frac13-1 && (textmore pedantic arithmetic) \
&= frac41 + frac12 && (textand some more) \
&= frac8+12 && (textalmost there) \
&= frac92. && (textdone!)
endalign
answered Jan 25 '18 at 16:02
Xander HendersonXander Henderson
14.9k103555
$endgroup$
$begingroup$
IT looks very understandable now, I jotted down my own version on a piece of paper here at my desk. My worries and further question is more about the justification of splitting the Series into two separate series... Doesn't there have to be the assumption such that the combined original series has to be convergent to begin with before manipulating? $ sum_n=1^infty frac2^n+1+13^n$
$endgroup$
– Late347
Jan 25 '18 at 16:31
$begingroup$
EDIT I just wanted to ask for sure, because I don't want to do silly mistakes with the infinite series... I still have a bad taste in my own mouth after having watched the numberphile video about summing divergent infinite series into each other (the video can be found on youtube with "1+2+3... = -1/12")
$endgroup$
– Late347
Jan 25 '18 at 16:33
$begingroup$
The justification kind of comes in through the back door. If $sum a_n$ and $sum b_n$ are convergent, then $sum (a_n + b_n)$ converges, and $sum (a_n + b_n) = sum a_n + sum b_n$. In the argument above, we break up a series of the form $sum (a_n + b_n)$ into two different convergent series $sum a_n$ and $sum b_n$, then use the fact that both of these converge in order to show that the original series converges.
$endgroup$
– Xander Henderson
Jan 25 '18 at 16:39
$begingroup$
On the other hand, if all you need to know is that the original series converges (and this might assuage your conscience regarding this assumption), you might note that $2^n+1 + 1 < 2^n+2$ for all $n$, and $sum frac2^n+23^n$ is a convergent geometric series. But then $$ 0 < sum frac2^n+1+13^n le sum frac2^n+23^n < infty, $$ which demonstrates that the original series does, in fact, converge.
$endgroup$
– Xander Henderson
Jan 25 '18 at 16:43
add a comment |
$begingroup$
After 20 years, I still have to rederive the identity
$$ sum_n=1^infty alpha^n = fracalpha1-alpha tag1$$
whenever I need it. Since it isn't that difficult, perhaps we should do that first.
Suppose that $|alpha| < 1$, and let
$$ S_k := sum_n=1^k alpha^k = alpha + alpha^2 + dotsb + alpha^k $$
be the $k$-th partial sum. Then
$$ alpha S_k = sum_n=1^k alpha^k+1 = alpha^2 + alpha^3 + dotsb + alpha^k+1. $$
Subtracting, we obtain
beginalign
(1-alpha) S_k
&= S_k - alpha S_k \
&= left( alpha + alpha^2 + dotsb + alpha^k right) - left( alpha^2 + alpha^3 + dotsb + alpha^k+1 right) \
&= alpha + left( alpha^2 - alpha^2right) + left( alpha^3 - alpha^3right) + dotsb + left(alpha^k - alpha^kright) - alpha^k+1 \
&= alpha - alpha^k+1.
endalign
Solving for $S_k$, we get
$$ S_k = fracalpha - alpha^k+11-alpha. $$
Since $|alpha| < 1$, it follows that $lim_ktoinfty alpha^k+1 = 0$, and so
$$
sum_n=1^infty alpha^n
:= lim_kto infty sum_n=1^k alpha^n
= lim_ktoinfty fracalpha - alpha^k+11-alpha
= fracalpha1-alpha.
$$
But this is exactly the identity at (1). Huzzah!
Now, how do we use this here?
Observe that
$$ frac2^n+1 + 13^n
= frac2^n+13^n + frac13^n
= 2 left( frac23 right)^n + left( frac13 right)^n.
tag2$$
But series are linear, and so play nice with addition and multiplication. Specifically, if $sum a_n$ and $sum b_n$ converge, and $C$ is any constant, then we have
$$ sum_n=1^infty left[ a_n + b_n right] = sum_n=1^infty a_n + sum_n=1^infty b_n
qquadtextandqquad
sum_n=1^infty Ca_n = C sum_n=1^infty a_n. $$
These are basic rules for "simplifying" series, which are going to be quite useful for you. If you are familiar with integration, you can perform exactly these manipulations if you replace the $sum$ with $int$, and the sequences with functions. If you are not familiar with integration, ignore this comment for now, but maybe keep it in the back of your mind for later.
beginalign
sum_n=1^infty frac2^n+1 + 13^n
&= sum_n=1^infty left[ 2 left( frac23 right)^n + left( frac13 right)^n right] && (textby application of (2)) \
&= 2sum_n=1^infty left( frac23 right)^n + sum_n=1^infty left( frac13 right)^n && (textsimplify the series) \
&= 2 cdot fracfrac231-frac23 + fracfrac131- frac13 && (textby application of (1)) \
&= 2 cdot fracfrac231-frac23cdot frac33 + fracfrac131- frac13cdot frac33 && (textsilly, pedantic arithmetic) \
&= 2 cdot frac23-2 + frac13-1 && (textmore pedantic arithmetic) \
&= frac41 + frac12 && (textand some more) \
&= frac8+12 && (textalmost there) \
&= frac92. && (textdone!)
endalign
answered Jan 25 '18 at 16:02
Xander HendersonXander Henderson
14.9k103555
$endgroup$
$begingroup$
IT looks very understandable now, I jotted down my own version on a piece of paper here at my desk. My worries and further question is more about the justification of splitting the Series into two separate series... Doesn't there have to be the assumption such that the combined original series has to be convergent to begin with before manipulating? $ sum_n=1^infty frac2^n+1+13^n$
$endgroup$
– Late347
Jan 25 '18 at 16:31
$begingroup$
EDIT I just wanted to ask for sure, because I don't want to do silly mistakes with the infinite series... I still have a bad taste in my own mouth after having watched the numberphile video about summing divergent infinite series into each other (the video can be found on youtube with "1+2+3... = -1/12")
$endgroup$
– Late347
Jan 25 '18 at 16:33
$begingroup$
The justification kind of comes in through the back door. If $sum a_n$ and $sum b_n$ are convergent, then $sum (a_n + b_n)$ converges, and $sum (a_n + b_n) = sum a_n + sum b_n$. In the argument above, we break up a series of the form $sum (a_n + b_n)$ into two different convergent series $sum a_n$ and $sum b_n$, then use the fact that both of these converge in order to show that the original series converges.
$endgroup$
– Xander Henderson
Jan 25 '18 at 16:39
$begingroup$
On the other hand, if all you need to know is that the original series converges (and this might assuage your conscience regarding this assumption), you might note that $2^n+1 + 1 < 2^n+2$ for all $n$, and $sum frac2^n+23^n$ is a convergent geometric series. But then $$ 0 < sum frac2^n+1+13^n le sum frac2^n+23^n < infty, $$ which demonstrates that the original series does, in fact, converge.
$endgroup$
– Xander Henderson
Jan 25 '18 at 16:43
add a comment |
$begingroup$
After 20 years, I still have to rederive the identity
$$ sum_n=1^infty alpha^n = fracalpha1-alpha tag1$$
whenever I need it. Since it isn't that difficult, perhaps we should do that first.
Suppose that $|alpha| < 1$, and let
$$ S_k := sum_n=1^k alpha^k = alpha + alpha^2 + dotsb + alpha^k $$
be the $k$-th partial sum. Then
$$ alpha S_k = sum_n=1^k alpha^k+1 = alpha^2 + alpha^3 + dotsb + alpha^k+1. $$
Subtracting, we obtain
beginalign
(1-alpha) S_k
&= S_k - alpha S_k \
&= left( alpha + alpha^2 + dotsb + alpha^k right) - left( alpha^2 + alpha^3 + dotsb + alpha^k+1 right) \
&= alpha + left( alpha^2 - alpha^2right) + left( alpha^3 - alpha^3right) + dotsb + left(alpha^k - alpha^kright) - alpha^k+1 \
&= alpha - alpha^k+1.
endalign
Solving for $S_k$, we get
$$ S_k = fracalpha - alpha^k+11-alpha. $$
Since $|alpha| < 1$, it follows that $lim_ktoinfty alpha^k+1 = 0$, and so
$$
sum_n=1^infty alpha^n
:= lim_kto infty sum_n=1^k alpha^n
= lim_ktoinfty fracalpha - alpha^k+11-alpha
= fracalpha1-alpha.
$$
But this is exactly the identity at (1). Huzzah!
Now, how do we use this here?
Observe that
$$ frac2^n+1 + 13^n
= frac2^n+13^n + frac13^n
= 2 left( frac23 right)^n + left( frac13 right)^n.
tag2$$
But series are linear, and so play nice with addition and multiplication. Specifically, if $sum a_n$ and $sum b_n$ converge, and $C$ is any constant, then we have
$$ sum_n=1^infty left[ a_n + b_n right] = sum_n=1^infty a_n + sum_n=1^infty b_n
qquadtextandqquad
sum_n=1^infty Ca_n = C sum_n=1^infty a_n. $$
These are basic rules for "simplifying" series, which are going to be quite useful for you. If you are familiar with integration, you can perform exactly these manipulations if you replace the $sum$ with $int$, and the sequences with functions. If you are not familiar with integration, ignore this comment for now, but maybe keep it in the back of your mind for later.
beginalign
sum_n=1^infty frac2^n+1 + 13^n
&= sum_n=1^infty left[ 2 left( frac23 right)^n + left( frac13 right)^n right] && (textby application of (2)) \
&= 2sum_n=1^infty left( frac23 right)^n + sum_n=1^infty left( frac13 right)^n && (textsimplify the series) \
&= 2 cdot fracfrac231-frac23 + fracfrac131- frac13 && (textby application of (1)) \
&= 2 cdot fracfrac231-frac23cdot frac33 + fracfrac131- frac13cdot frac33 && (textsilly, pedantic arithmetic) \
&= 2 cdot frac23-2 + frac13-1 && (textmore pedantic arithmetic) \
&= frac41 + frac12 && (textand some more) \
&= frac8+12 && (textalmost there) \
&= frac92. && (textdone!)
endalign
answered Jan 25 '18 at 16:02
Xander HendersonXander Henderson
14.9k103555
$endgroup$
After 20 years, I still have to rederive the identity
$$ sum_n=1^infty alpha^n = fracalpha1-alpha tag1$$
whenever I need it. Since it isn't that difficult, perhaps we should do that first.
Suppose that $|alpha| < 1$, and let
$$ S_k := sum_n=1^k alpha^k = alpha + alpha^2 + dotsb + alpha^k $$
be the $k$-th partial sum. Then
$$ alpha S_k = sum_n=1^k alpha^k+1 = alpha^2 + alpha^3 + dotsb + alpha^k+1. $$
Subtracting, we obtain
beginalign
(1-alpha) S_k
&= S_k - alpha S_k \
&= left( alpha + alpha^2 + dotsb + alpha^k right) - left( alpha^2 + alpha^3 + dotsb + alpha^k+1 right) \
&= alpha + left( alpha^2 - alpha^2right) + left( alpha^3 - alpha^3right) + dotsb + left(alpha^k - alpha^kright) - alpha^k+1 \
&= alpha - alpha^k+1.
endalign
Solving for $S_k$, we get
$$ S_k = fracalpha - alpha^k+11-alpha. $$
Since $|alpha| < 1$, it follows that $lim_ktoinfty alpha^k+1 = 0$, and so
$$
sum_n=1^infty alpha^n
:= lim_kto infty sum_n=1^k alpha^n
= lim_ktoinfty fracalpha - alpha^k+11-alpha
= fracalpha1-alpha.
$$
But this is exactly the identity at (1). Huzzah!
Now, how do we use this here?
Observe that
$$ frac2^n+1 + 13^n
= frac2^n+13^n + frac13^n
= 2 left( frac23 right)^n + left( frac13 right)^n.
tag2$$
But series are linear, and so play nice with addition and multiplication. Specifically, if $sum a_n$ and $sum b_n$ converge, and $C$ is any constant, then we have
$$ sum_n=1^infty left[ a_n + b_n right] = sum_n=1^infty a_n + sum_n=1^infty b_n
qquadtextandqquad
sum_n=1^infty Ca_n = C sum_n=1^infty a_n. $$
These are basic rules for "simplifying" series, which are going to be quite useful for you. If you are familiar with integration, you can perform exactly these manipulations if you replace the $sum$ with $int$, and the sequences with functions. If you are not familiar with integration, ignore this comment for now, but maybe keep it in the back of your mind for later.
beginalign
sum_n=1^infty frac2^n+1 + 13^n
&= sum_n=1^infty left[ 2 left( frac23 right)^n + left( frac13 right)^n right] && (textby application of (2)) \
&= 2sum_n=1^infty left( frac23 right)^n + sum_n=1^infty left( frac13 right)^n && (textsimplify the series) \
&= 2 cdot fracfrac231-frac23 + fracfrac131- frac13 && (textby application of (1)) \
&= 2 cdot fracfrac231-frac23cdot frac33 + fracfrac131- frac13cdot frac33 && (textsilly, pedantic arithmetic) \
&= 2 cdot frac23-2 + frac13-1 && (textmore pedantic arithmetic) \
&= frac41 + frac12 && (textand some more) \
&= frac8+12 && (textalmost there) \
&= frac92. && (textdone!)
endalign
answered Jan 25 '18 at 16:02
Xander HendersonXander Henderson
14.9k103555
edited Jan 25 '18 at 16:18
answered Jan 25 '18 at 16:02
Xander HendersonXander Henderson
14.9k103555
answered Jan 25 '18 at 16:02
Xander HendersonXander Henderson
14.9k103555
answered Jan 25 '18 at 16:02
Xander HendersonXander Henderson
14.9k103555
14.9k103555
$begingroup$
IT looks very understandable now, I jotted down my own version on a piece of paper here at my desk. My worries and further question is more about the justification of splitting the Series into two separate series... Doesn't there have to be the assumption such that the combined original series has to be convergent to begin with before manipulating? $ sum_n=1^infty frac2^n+1+13^n$
$endgroup$
– Late347
Jan 25 '18 at 16:31
$begingroup$
EDIT I just wanted to ask for sure, because I don't want to do silly mistakes with the infinite series... I still have a bad taste in my own mouth after having watched the numberphile video about summing divergent infinite series into each other (the video can be found on youtube with "1+2+3... = -1/12")
$endgroup$
– Late347
Jan 25 '18 at 16:33
$begingroup$
The justification kind of comes in through the back door. If $sum a_n$ and $sum b_n$ are convergent, then $sum (a_n + b_n)$ converges, and $sum (a_n + b_n) = sum a_n + sum b_n$. In the argument above, we break up a series of the form $sum (a_n + b_n)$ into two different convergent series $sum a_n$ and $sum b_n$, then use the fact that both of these converge in order to show that the original series converges.
$endgroup$
– Xander Henderson
Jan 25 '18 at 16:39
$begingroup$
On the other hand, if all you need to know is that the original series converges (and this might assuage your conscience regarding this assumption), you might note that $2^n+1 + 1 < 2^n+2$ for all $n$, and $sum frac2^n+23^n$ is a convergent geometric series. But then $$ 0 < sum frac2^n+1+13^n le sum frac2^n+23^n < infty, $$ which demonstrates that the original series does, in fact, converge.
$endgroup$
– Xander Henderson
Jan 25 '18 at 16:43
add a comment |
$begingroup$
IT looks very understandable now, I jotted down my own version on a piece of paper here at my desk. My worries and further question is more about the justification of splitting the Series into two separate series... Doesn't there have to be the assumption such that the combined original series has to be convergent to begin with before manipulating? $ sum_n=1^infty frac2^n+1+13^n$
$endgroup$
– Late347
Jan 25 '18 at 16:31
$begingroup$
EDIT I just wanted to ask for sure, because I don't want to do silly mistakes with the infinite series... I still have a bad taste in my own mouth after having watched the numberphile video about summing divergent infinite series into each other (the video can be found on youtube with "1+2+3... = -1/12")
$endgroup$
– Late347
Jan 25 '18 at 16:33
$begingroup$
The justification kind of comes in through the back door. If $sum a_n$ and $sum b_n$ are convergent, then $sum (a_n + b_n)$ converges, and $sum (a_n + b_n) = sum a_n + sum b_n$. In the argument above, we break up a series of the form $sum (a_n + b_n)$ into two different convergent series $sum a_n$ and $sum b_n$, then use the fact that both of these converge in order to show that the original series converges.
$endgroup$
– Xander Henderson
Jan 25 '18 at 16:39
$begingroup$
On the other hand, if all you need to know is that the original series converges (and this might assuage your conscience regarding this assumption), you might note that $2^n+1 + 1 < 2^n+2$ for all $n$, and $sum frac2^n+23^n$ is a convergent geometric series. But then $$ 0 < sum frac2^n+1+13^n le sum frac2^n+23^n < infty, $$ which demonstrates that the original series does, in fact, converge.
$endgroup$
– Xander Henderson
Jan 25 '18 at 16:43
$begingroup$
IT looks very understandable now, I jotted down my own version on a piece of paper here at my desk. My worries and further question is more about the justification of splitting the Series into two separate series... Doesn't there have to be the assumption such that the combined original series has to be convergent to begin with before manipulating? $ sum_n=1^infty frac2^n+1+13^n$
$endgroup$
– Late347
Jan 25 '18 at 16:31
$begingroup$
IT looks very understandable now, I jotted down my own version on a piece of paper here at my desk. My worries and further question is more about the justification of splitting the Series into two separate series... Doesn't there have to be the assumption such that the combined original series has to be convergent to begin with before manipulating? $ sum_n=1^infty frac2^n+1+13^n$
$endgroup$
– Late347
Jan 25 '18 at 16:31
$begingroup$
EDIT I just wanted to ask for sure, because I don't want to do silly mistakes with the infinite series... I still have a bad taste in my own mouth after having watched the numberphile video about summing divergent infinite series into each other (the video can be found on youtube with "1+2+3... = -1/12")
$endgroup$
– Late347
Jan 25 '18 at 16:33
$begingroup$
EDIT I just wanted to ask for sure, because I don't want to do silly mistakes with the infinite series... I still have a bad taste in my own mouth after having watched the numberphile video about summing divergent infinite series into each other (the video can be found on youtube with "1+2+3... = -1/12")
$endgroup$
– Late347
Jan 25 '18 at 16:33
$begingroup$
The justification kind of comes in through the back door. If $sum a_n$ and $sum b_n$ are convergent, then $sum (a_n + b_n)$ converges, and $sum (a_n + b_n) = sum a_n + sum b_n$. In the argument above, we break up a series of the form $sum (a_n + b_n)$ into two different convergent series $sum a_n$ and $sum b_n$, then use the fact that both of these converge in order to show that the original series converges.
$endgroup$
– Xander Henderson
Jan 25 '18 at 16:39
$begingroup$
The justification kind of comes in through the back door. If $sum a_n$ and $sum b_n$ are convergent, then $sum (a_n + b_n)$ converges, and $sum (a_n + b_n) = sum a_n + sum b_n$. In the argument above, we break up a series of the form $sum (a_n + b_n)$ into two different convergent series $sum a_n$ and $sum b_n$, then use the fact that both of these converge in order to show that the original series converges.
$endgroup$
– Xander Henderson
Jan 25 '18 at 16:39
$begingroup$
On the other hand, if all you need to know is that the original series converges (and this might assuage your conscience regarding this assumption), you might note that $2^n+1 + 1 < 2^n+2$ for all $n$, and $sum frac2^n+23^n$ is a convergent geometric series. But then $$ 0 < sum frac2^n+1+13^n le sum frac2^n+23^n < infty, $$ which demonstrates that the original series does, in fact, converge.
$endgroup$
– Xander Henderson
Jan 25 '18 at 16:43
$begingroup$
On the other hand, if all you need to know is that the original series converges (and this might assuage your conscience regarding this assumption), you might note that $2^n+1 + 1 < 2^n+2$ for all $n$, and $sum frac2^n+23^n$ is a convergent geometric series. But then $$ 0 < sum frac2^n+1+13^n le sum frac2^n+23^n < infty, $$ which demonstrates that the original series does, in fact, converge.
$endgroup$
– Xander Henderson
Jan 25 '18 at 16:43
add a comment |
$begingroup$
You can split the sum and use
$$sum_n=1^infty r^n=fracr1-r$$
answered Jan 25 '18 at 15:46
gimusigimusi
93.1k84594
$endgroup$
add a comment |
$begingroup$
You can split the sum and use
$$sum_n=1^infty r^n=fracr1-r$$
answered Jan 25 '18 at 15:46
gimusigimusi
93.1k84594
$endgroup$
add a comment |
$begingroup$
You can split the sum and use
$$sum_n=1^infty r^n=fracr1-r$$
answered Jan 25 '18 at 15:46
gimusigimusi
93.1k84594
$endgroup$
You can split the sum and use
$$sum_n=1^infty r^n=fracr1-r$$
answered Jan 25 '18 at 15:46
gimusigimusi
93.1k84594
answered Jan 25 '18 at 15:46
gimusigimusi
93.1k84594
answered Jan 25 '18 at 15:46
gimusigimusi
93.1k84594
answered Jan 25 '18 at 15:46
gimusigimusi
93.1k84594
93.1k84594
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2620771%2fhelp-with-sum-of-infinite-series-stuck-in-problem%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
