Proving an alternating infinite series to be divergent The Next CEO of Stack OverflowIs there an alternating series that satisfies only one of the conditions of the Alternating Series Test that nonetheless converges?Determine whether $sum_n=1^+inftyc_n$ for a given $c_n$ is divergent or convergentDivergent Alternating SeriesDivergent Infinite Series QuestionHow can I find the sum of an infinite series of products?Sum of an infinite series of fractionsDivergent alternating series with negative first derivative?Divergent Series Finite SumAlternating combination of a convergent and a divergent seriesconvergence of infinite series $sumlimits_n=0^inftyfrac(1+1/2+1/3+…+1/n)n$
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Proving an alternating infinite series to be divergent
The Next CEO of Stack OverflowIs there an alternating series that satisfies only one of the conditions of the Alternating Series Test that nonetheless converges?Determine whether $sum_n=1^+inftyc_n$ for a given $c_n$ is divergent or convergentDivergent Alternating SeriesDivergent Infinite Series QuestionHow can I find the sum of an infinite series of products?Sum of an infinite series of fractionsDivergent alternating series with negative first derivative?Divergent Series Finite SumAlternating combination of a convergent and a divergent seriesconvergence of infinite series $sumlimits_n=0^inftyfrac(1+1/2+1/3+…+1/n)n$
$begingroup$
I am trying to prove $$sum_n=0^infty frac(-4)^3n5^n-1$$ is a divergent series. I know that it increases as the series progresses regardless of sign, so it must be divergent, but im not sure how to prove it.
Usually with an alternating infinite series as long as the term not involving the minus doesnt tend to zero then it is easy to prove, however in this, that term is $$frac15^n-1$$ (I think), so it makes it a little harder, maybe i'm taking the wrong appraoch to this, any help?
sequences-and-series convergence divergent-series
edited yesterday
José Carlos Santos
171k23132240
asked yesterday
JackJack
63
New contributor
Jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am trying to prove $$sum_n=0^infty frac(-4)^3n5^n-1$$ is a divergent series. I know that it increases as the series progresses regardless of sign, so it must be divergent, but im not sure how to prove it.
Usually with an alternating infinite series as long as the term not involving the minus doesnt tend to zero then it is easy to prove, however in this, that term is $$frac15^n-1$$ (I think), so it makes it a little harder, maybe i'm taking the wrong appraoch to this, any help?
sequences-and-series convergence divergent-series
edited yesterday
José Carlos Santos
171k23132240
asked yesterday
JackJack
63
New contributor
Jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Hint: what is $left|fraca_n+1a_nright|$? Or, write the summand with an explicit form of $ar^n$, to make it look clearly like a geometric series, then check that $|r|ge 1$ (which implies that the geometric series is divergent, e.g. since the $n$-th term will not tend to $0$ as $ntoinfty$).
$endgroup$
– Minus One-Twelfth
yesterday
add a comment |
$begingroup$
I am trying to prove $$sum_n=0^infty frac(-4)^3n5^n-1$$ is a divergent series. I know that it increases as the series progresses regardless of sign, so it must be divergent, but im not sure how to prove it.
Usually with an alternating infinite series as long as the term not involving the minus doesnt tend to zero then it is easy to prove, however in this, that term is $$frac15^n-1$$ (I think), so it makes it a little harder, maybe i'm taking the wrong appraoch to this, any help?
sequences-and-series convergence divergent-series
edited yesterday
José Carlos Santos
171k23132240
asked yesterday
JackJack
63
New contributor
Jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am trying to prove $$sum_n=0^infty frac(-4)^3n5^n-1$$ is a divergent series. I know that it increases as the series progresses regardless of sign, so it must be divergent, but im not sure how to prove it.
Usually with an alternating infinite series as long as the term not involving the minus doesnt tend to zero then it is easy to prove, however in this, that term is $$frac15^n-1$$ (I think), so it makes it a little harder, maybe i'm taking the wrong appraoch to this, any help?
sequences-and-series convergence divergent-series
sequences-and-series convergence divergent-series
edited yesterday
José Carlos Santos
171k23132240
asked yesterday
JackJack
63
New contributor
Jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
José Carlos Santos
171k23132240
asked yesterday
JackJack
63
New contributor
Jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
José Carlos Santos
171k23132240
edited yesterday
José Carlos Santos
171k23132240
edited yesterday
José Carlos Santos
171k23132240
171k23132240
asked yesterday
JackJack
63
New contributor
Jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
JackJack
63
asked yesterday
JackJack
63
63
New contributor
Jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Hint: what is $left|fraca_n+1a_nright|$? Or, write the summand with an explicit form of $ar^n$, to make it look clearly like a geometric series, then check that $|r|ge 1$ (which implies that the geometric series is divergent, e.g. since the $n$-th term will not tend to $0$ as $ntoinfty$).
$endgroup$
– Minus One-Twelfth
yesterday
add a comment |
$begingroup$
Hint: what is $left|fraca_n+1a_nright|$? Or, write the summand with an explicit form of $ar^n$, to make it look clearly like a geometric series, then check that $|r|ge 1$ (which implies that the geometric series is divergent, e.g. since the $n$-th term will not tend to $0$ as $ntoinfty$).
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
Hint: what is $left|fraca_n+1a_nright|$? Or, write the summand with an explicit form of $ar^n$, to make it look clearly like a geometric series, then check that $|r|ge 1$ (which implies that the geometric series is divergent, e.g. since the $n$-th term will not tend to $0$ as $ntoinfty$).
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
Hint: what is $left|fraca_n+1a_nright|$? Or, write the summand with an explicit form of $ar^n$, to make it look clearly like a geometric series, then check that $|r|ge 1$ (which implies that the geometric series is divergent, e.g. since the $n$-th term will not tend to $0$ as $ntoinfty$).
$endgroup$
– Minus One-Twelfth
yesterday
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint:
$$dfrac(-4)^3n5^n-1=5left(dfrac-645right)^n$$
Use this
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158278
$endgroup$
$begingroup$
Wow that makes it alot easier! Thank you! Sorry I am new to the topic.
$endgroup$
– Jack
yesterday
add a comment |
$begingroup$
Since $lim_ntoinftyleftlvertfrac(-4)^3n5^n-1rightrvert=infty$, you don't have $lim_ntoinftyfrac(-4)^3n5^n-1=0$, and therefore the series diverges.
answered yesterday
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Hint
Use the d'alembert Ratio Test.
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
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$begingroup$
Hint:
$$dfrac(-4)^3n5^n-1=5left(dfrac-645right)^n$$
Use this
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158278
$endgroup$
$begingroup$
Wow that makes it alot easier! Thank you! Sorry I am new to the topic.
$endgroup$
– Jack
yesterday
add a comment |
$begingroup$
Hint:
$$dfrac(-4)^3n5^n-1=5left(dfrac-645right)^n$$
Use this
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158278
$endgroup$
$begingroup$
Wow that makes it alot easier! Thank you! Sorry I am new to the topic.
$endgroup$
– Jack
yesterday
add a comment |
$begingroup$
Hint:
$$dfrac(-4)^3n5^n-1=5left(dfrac-645right)^n$$
Use this
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158278
$endgroup$
Hint:
$$dfrac(-4)^3n5^n-1=5left(dfrac-645right)^n$$
Use this
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158278
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158278
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158278
answered yesterday
lab bhattacharjeelab bhattacharjee
228k15158278
228k15158278
$begingroup$
Wow that makes it alot easier! Thank you! Sorry I am new to the topic.
$endgroup$
– Jack
yesterday
add a comment |
$begingroup$
Wow that makes it alot easier! Thank you! Sorry I am new to the topic.
$endgroup$
– Jack
yesterday
$begingroup$
Wow that makes it alot easier! Thank you! Sorry I am new to the topic.
$endgroup$
– Jack
yesterday
$begingroup$
Wow that makes it alot easier! Thank you! Sorry I am new to the topic.
$endgroup$
– Jack
yesterday
add a comment |
$begingroup$
Since $lim_ntoinftyleftlvertfrac(-4)^3n5^n-1rightrvert=infty$, you don't have $lim_ntoinftyfrac(-4)^3n5^n-1=0$, and therefore the series diverges.
answered yesterday
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Since $lim_ntoinftyleftlvertfrac(-4)^3n5^n-1rightrvert=infty$, you don't have $lim_ntoinftyfrac(-4)^3n5^n-1=0$, and therefore the series diverges.
answered yesterday
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Since $lim_ntoinftyleftlvertfrac(-4)^3n5^n-1rightrvert=infty$, you don't have $lim_ntoinftyfrac(-4)^3n5^n-1=0$, and therefore the series diverges.
answered yesterday
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
Since $lim_ntoinftyleftlvertfrac(-4)^3n5^n-1rightrvert=infty$, you don't have $lim_ntoinftyfrac(-4)^3n5^n-1=0$, and therefore the series diverges.
answered yesterday
José Carlos SantosJosé Carlos Santos
171k23132240
edited yesterday
answered yesterday
José Carlos SantosJosé Carlos Santos
171k23132240
answered yesterday
José Carlos SantosJosé Carlos Santos
171k23132240
answered yesterday
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
add a comment |
add a comment |
$begingroup$
Hint
Use the d'alembert Ratio Test.
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Hint
Use the d'alembert Ratio Test.
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Hint
Use the d'alembert Ratio Test.
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
Hint
Use the d'alembert Ratio Test.
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
add a comment |
add a comment |
Jack is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Hint: what is $left|fraca_n+1a_nright|$? Or, write the summand with an explicit form of $ar^n$, to make it look clearly like a geometric series, then check that $|r|ge 1$ (which implies that the geometric series is divergent, e.g. since the $n$-th term will not tend to $0$ as $ntoinfty$).
$endgroup$
– Minus One-Twelfth
yesterday