Show if $a_n^2$ is bounded then $a_n$ is boundedIf $b_n$ is a bounded sequence and $lim a_n = 0$, show that $lim(a_nb_n) = 0$$b_n$ bounded, $sum a_n$ converges absolutely, then $sum a_nb_n$ alsoShow that a sequence is bounded if and only if there exists a K $inmathbbR$ such that $| a_n | leq K$ $forall nin mathbbN$.$a_n$ is bounded and decreasingBounded convex sequenceQuestion about post http://math.stackexchange.com/questions/1568696/lim-n-rightarrow-infty-a-n-a-how-to-prove-lim-n-rightarrow-infty/1568740#1568740Bounded sequences.Suppose $(a_n)$ is a sequence in $mathbbR$ such that $a_n leq b forall n$ and $a_n rightarrow a$, same $a$, then $a leq b$If $a_n$ is bounded and non-decreasing, prove that $liminf b_n = 0$, $b_n = n(a_n+1 - a_n)$If the limit exists and is smaller than 1 then $limsup sqrt[n]a_n leq lim_nrightarrowinfty |fraca_n+1a_n|$
What is the command to reset a PC without deleting any files
Landing in very high winds
What does 'script /dev/null' do?
Why is my log file so massive? 22gb. I am running log backups
What does "enim et" mean?
Could Giant Ground Sloths have been a good pack animal for the ancient Mayans?
Patience, young "Padovan"
If a centaur druid Wild Shapes into a Giant Elk, do their Charge features stack?
What to wear for invited talk in Canada
How could a lack of term limits lead to a "dictatorship?"
Can I legally use front facing blue light in the UK?
aging parents with no investments
Are objects structures and/or vice versa?
Why do UK politicians seemingly ignore opinion polls on Brexit?
Why airport relocation isn't done gradually?
How to answer pointed "are you quitting" questioning when I don't want them to suspect
Symmetry in quantum mechanics
Where to refill my bottle in India?
Need help identifying/translating a plaque in Tangier, Morocco
When blogging recipes, how can I support both readers who want the narrative/journey and ones who want the printer-friendly recipe?
I see my dog run
Is Social Media Science Fiction?
Is it legal to have the "// (c) 2019 John Smith" header in all files when there are hundreds of contributors?
Is this food a bread or a loaf?
Show if $a_n^2$ is bounded then $a_n$ is bounded
If $b_n$ is a bounded sequence and $lim a_n = 0$, show that $lim(a_nb_n) = 0$$b_n$ bounded, $sum a_n$ converges absolutely, then $sum a_nb_n$ alsoShow that a sequence is bounded if and only if there exists a K $inmathbbR$ such that $| a_n | leq K$ $forall nin mathbbN$.$a_n$ is bounded and decreasingBounded convex sequenceQuestion about post http://math.stackexchange.com/questions/1568696/lim-n-rightarrow-infty-a-n-a-how-to-prove-lim-n-rightarrow-infty/1568740#1568740Bounded sequences.Suppose $(a_n)$ is a sequence in $mathbbR$ such that $a_n leq b forall n$ and $a_n rightarrow a$, same $a$, then $a leq b$If $a_n$ is bounded and non-decreasing, prove that $liminf b_n = 0$, $b_n = n(a_n+1 - a_n)$If the limit exists and is smaller than 1 then $limsup sqrt[n]a_n leq lim_nrightarrowinfty |fraca_n+1a_n|$
$begingroup$
If $a_n^2$ is bounded then I know for a non-negative number $M$, $mid a_n^2 mid leq M, forall n$.
How do I show that this leads to $a_n$ being bounded?
$mid a_n^2 mid leq M$
$-M leq a_n^2 leq M$
but now I don't think I can replace $M$ with $sqrtM$, can I?
$-sqrtM leq a_n leq sqrtM$ is probably allowed, but how do I justify this?
real-analysis inequality
asked Mar 30 at 4:23
SunnySunny
788
$endgroup$
add a comment |
$begingroup$
If $a_n^2$ is bounded then I know for a non-negative number $M$, $mid a_n^2 mid leq M, forall n$.
How do I show that this leads to $a_n$ being bounded?
$mid a_n^2 mid leq M$
$-M leq a_n^2 leq M$
but now I don't think I can replace $M$ with $sqrtM$, can I?
$-sqrtM leq a_n leq sqrtM$ is probably allowed, but how do I justify this?
real-analysis inequality
asked Mar 30 at 4:23
SunnySunny
788
$endgroup$
1
$begingroup$
Letting $sqrt$ denote the positive branch of the square root, your justification is correct: $$|a_n| leq sqrtM.$$
$endgroup$
– avs
Mar 30 at 4:26
$begingroup$
Is $a_n$ restricted to be real?
$endgroup$
– L. F.
Mar 30 at 4:26
$begingroup$
Yes, $a_n in mathbbR$
$endgroup$
– Sunny
Mar 30 at 4:28
add a comment |
$begingroup$
If $a_n^2$ is bounded then I know for a non-negative number $M$, $mid a_n^2 mid leq M, forall n$.
How do I show that this leads to $a_n$ being bounded?
$mid a_n^2 mid leq M$
$-M leq a_n^2 leq M$
but now I don't think I can replace $M$ with $sqrtM$, can I?
$-sqrtM leq a_n leq sqrtM$ is probably allowed, but how do I justify this?
real-analysis inequality
asked Mar 30 at 4:23
SunnySunny
788
$endgroup$
If $a_n^2$ is bounded then I know for a non-negative number $M$, $mid a_n^2 mid leq M, forall n$.
How do I show that this leads to $a_n$ being bounded?
$mid a_n^2 mid leq M$
$-M leq a_n^2 leq M$
but now I don't think I can replace $M$ with $sqrtM$, can I?
$-sqrtM leq a_n leq sqrtM$ is probably allowed, but how do I justify this?
real-analysis inequality
real-analysis inequality
asked Mar 30 at 4:23
SunnySunny
788
asked Mar 30 at 4:23
SunnySunny
788
asked Mar 30 at 4:23
SunnySunny
788
asked Mar 30 at 4:23
SunnySunny
788
asked Mar 30 at 4:23
SunnySunny
788
788
1
$begingroup$
Letting $sqrt$ denote the positive branch of the square root, your justification is correct: $$|a_n| leq sqrtM.$$
$endgroup$
– avs
Mar 30 at 4:26
$begingroup$
Is $a_n$ restricted to be real?
$endgroup$
– L. F.
Mar 30 at 4:26
$begingroup$
Yes, $a_n in mathbbR$
$endgroup$
– Sunny
Mar 30 at 4:28
add a comment |
1
$begingroup$
Letting $sqrt$ denote the positive branch of the square root, your justification is correct: $$|a_n| leq sqrtM.$$
$endgroup$
– avs
Mar 30 at 4:26
$begingroup$
Is $a_n$ restricted to be real?
$endgroup$
– L. F.
Mar 30 at 4:26
$begingroup$
Yes, $a_n in mathbbR$
$endgroup$
– Sunny
Mar 30 at 4:28
1
1
$begingroup$
Letting $sqrt$ denote the positive branch of the square root, your justification is correct: $$|a_n| leq sqrtM.$$
$endgroup$
– avs
Mar 30 at 4:26
$begingroup$
Letting $sqrt$ denote the positive branch of the square root, your justification is correct: $$|a_n| leq sqrtM.$$
$endgroup$
– avs
Mar 30 at 4:26
$begingroup$
Is $a_n$ restricted to be real?
$endgroup$
– L. F.
Mar 30 at 4:26
$begingroup$
Is $a_n$ restricted to be real?
$endgroup$
– L. F.
Mar 30 at 4:26
$begingroup$
Yes, $a_n in mathbbR$
$endgroup$
– Sunny
Mar 30 at 4:28
$begingroup$
Yes, $a_n in mathbbR$
$endgroup$
– Sunny
Mar 30 at 4:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The squaring function $f : [0,infty) to [0,infty)$, $f(x)=x^2$, is a strictly monotonic bijection, hence its inverse function $f^-1 : [0,infty) to [0,infty)$, $f^-1(x)=sqrtx$ is also a strictly monotonic bijection. So $|a^2_n| le M$ if and only if $|a_n| le sqrtM$. And that last inequality is equivalent to $-sqrtM le a_n le sqrtM$ simply by the definition of absolute values.
answered Mar 30 at 4:36
Lee MosherLee Mosher
51.8k33889
$endgroup$
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%2f3167926%2fshow-if-a-n2-is-bounded-then-a-n-is-bounded%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The squaring function $f : [0,infty) to [0,infty)$, $f(x)=x^2$, is a strictly monotonic bijection, hence its inverse function $f^-1 : [0,infty) to [0,infty)$, $f^-1(x)=sqrtx$ is also a strictly monotonic bijection. So $|a^2_n| le M$ if and only if $|a_n| le sqrtM$. And that last inequality is equivalent to $-sqrtM le a_n le sqrtM$ simply by the definition of absolute values.
answered Mar 30 at 4:36
Lee MosherLee Mosher
51.8k33889
$endgroup$
add a comment |
$begingroup$
The squaring function $f : [0,infty) to [0,infty)$, $f(x)=x^2$, is a strictly monotonic bijection, hence its inverse function $f^-1 : [0,infty) to [0,infty)$, $f^-1(x)=sqrtx$ is also a strictly monotonic bijection. So $|a^2_n| le M$ if and only if $|a_n| le sqrtM$. And that last inequality is equivalent to $-sqrtM le a_n le sqrtM$ simply by the definition of absolute values.
answered Mar 30 at 4:36
Lee MosherLee Mosher
51.8k33889
$endgroup$
add a comment |
$begingroup$
The squaring function $f : [0,infty) to [0,infty)$, $f(x)=x^2$, is a strictly monotonic bijection, hence its inverse function $f^-1 : [0,infty) to [0,infty)$, $f^-1(x)=sqrtx$ is also a strictly monotonic bijection. So $|a^2_n| le M$ if and only if $|a_n| le sqrtM$. And that last inequality is equivalent to $-sqrtM le a_n le sqrtM$ simply by the definition of absolute values.
answered Mar 30 at 4:36
Lee MosherLee Mosher
51.8k33889
$endgroup$
The squaring function $f : [0,infty) to [0,infty)$, $f(x)=x^2$, is a strictly monotonic bijection, hence its inverse function $f^-1 : [0,infty) to [0,infty)$, $f^-1(x)=sqrtx$ is also a strictly monotonic bijection. So $|a^2_n| le M$ if and only if $|a_n| le sqrtM$. And that last inequality is equivalent to $-sqrtM le a_n le sqrtM$ simply by the definition of absolute values.
answered Mar 30 at 4:36
Lee MosherLee Mosher
51.8k33889
answered Mar 30 at 4:36
Lee MosherLee Mosher
51.8k33889
answered Mar 30 at 4:36
Lee MosherLee Mosher
51.8k33889
answered Mar 30 at 4:36
Lee MosherLee Mosher
51.8k33889
51.8k33889
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%2f3167926%2fshow-if-a-n2-is-bounded-then-a-n-is-bounded%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
1
$begingroup$
Letting $sqrt$ denote the positive branch of the square root, your justification is correct: $$|a_n| leq sqrtM.$$
$endgroup$
– avs
Mar 30 at 4:26
$begingroup$
Is $a_n$ restricted to be real?
$endgroup$
– L. F.
Mar 30 at 4:26
$begingroup$
Yes, $a_n in mathbbR$
$endgroup$
– Sunny
Mar 30 at 4:28