Convergence of $a_k=sum_{k=5}^{+infty}(-1)^k({3over2})^{-k}(k^2+5)sin(k+5)$












0












$begingroup$


Firstly, $lim_{k to +infty}a_k=0$,so necessary condition for convergence is satisfied.If we start to study absolute convergence we have : $$a_k=({3over2})^{-k}(k^2+5)|sin(k+5)|leq({3over2})^{-k}(k^2+5)sim({3over2})^{-k}(k^2)=frac{k^2}{({3over2})^{k}}$$

So if we apply ratio test : $$lim_{kto+infty}frac{(k+1)^2({3over2})^k}{({3over2})^k({3over2})(k^2)}={2over3}lt1$$ And $a_k$ is absolutely and simply convergent.Is this approach correct?






calculus sequences-and-series power-series absolute-convergence







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Yes, this is correct. You could have used plain comparison, since $sin(k+5)$ is bounded and $(3/2)^{-k}$ goes to zero faster than, say, $k^{-4}$.
    $endgroup$
    – GReyes
    Jan 29 at 7:35










  • $begingroup$
    @GReyes what exactly our comparison function would look like?Probably ${k^2over k^4}={1over k^2}$?
    $endgroup$
    – Turan Nasibli
    Jan 29 at 7:37












  • $begingroup$
    I think your sum does converge.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 29 at 7:49










  • $begingroup$
    @Dr.SonnhardGraubner it does,for sure.
    $endgroup$
    – Turan Nasibli
    Jan 29 at 7:51










  • $begingroup$
    @TuranNəsibli Yes, you are right. For large enough $k$, $(3/2)^{-k}le k^{-4}$ and $sin(k+5)le 1$, therefore your $|a_k|le 1/k^2$.
    $endgroup$
    – GReyes
    Jan 29 at 20:04
















0












$begingroup$


Firstly, $lim_{k to +infty}a_k=0$,so necessary condition for convergence is satisfied.If we start to study absolute convergence we have : $$a_k=({3over2})^{-k}(k^2+5)|sin(k+5)|leq({3over2})^{-k}(k^2+5)sim({3over2})^{-k}(k^2)=frac{k^2}{({3over2})^{k}}$$

So if we apply ratio test : $$lim_{kto+infty}frac{(k+1)^2({3over2})^k}{({3over2})^k({3over2})(k^2)}={2over3}lt1$$ And $a_k$ is absolutely and simply convergent.Is this approach correct?






calculus sequences-and-series power-series absolute-convergence







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Yes, this is correct. You could have used plain comparison, since $sin(k+5)$ is bounded and $(3/2)^{-k}$ goes to zero faster than, say, $k^{-4}$.
    $endgroup$
    – GReyes
    Jan 29 at 7:35










  • $begingroup$
    @GReyes what exactly our comparison function would look like?Probably ${k^2over k^4}={1over k^2}$?
    $endgroup$
    – Turan Nasibli
    Jan 29 at 7:37












  • $begingroup$
    I think your sum does converge.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 29 at 7:49










  • $begingroup$
    @Dr.SonnhardGraubner it does,for sure.
    $endgroup$
    – Turan Nasibli
    Jan 29 at 7:51










  • $begingroup$
    @TuranNəsibli Yes, you are right. For large enough $k$, $(3/2)^{-k}le k^{-4}$ and $sin(k+5)le 1$, therefore your $|a_k|le 1/k^2$.
    $endgroup$
    – GReyes
    Jan 29 at 20:04














0












0








0





$begingroup$


Firstly, $lim_{k to +infty}a_k=0$,so necessary condition for convergence is satisfied.If we start to study absolute convergence we have : $$a_k=({3over2})^{-k}(k^2+5)|sin(k+5)|leq({3over2})^{-k}(k^2+5)sim({3over2})^{-k}(k^2)=frac{k^2}{({3over2})^{k}}$$

So if we apply ratio test : $$lim_{kto+infty}frac{(k+1)^2({3over2})^k}{({3over2})^k({3over2})(k^2)}={2over3}lt1$$ And $a_k$ is absolutely and simply convergent.Is this approach correct?






calculus sequences-and-series power-series absolute-convergence







share|cite|improve this question









$endgroup$




Firstly, $lim_{k to +infty}a_k=0$,so necessary condition for convergence is satisfied.If we start to study absolute convergence we have : $$a_k=({3over2})^{-k}(k^2+5)|sin(k+5)|leq({3over2})^{-k}(k^2+5)sim({3over2})^{-k}(k^2)=frac{k^2}{({3over2})^{k}}$$

So if we apply ratio test : $$lim_{kto+infty}frac{(k+1)^2({3over2})^k}{({3over2})^k({3over2})(k^2)}={2over3}lt1$$ And $a_k$ is absolutely and simply convergent.Is this approach correct?






calculus sequences-and-series power-series absolute-convergence




calculus sequences-and-series power-series absolute-convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 29 at 7:26









Turan NasibliTuran Nasibli

846




846








  • 2




    $begingroup$
    Yes, this is correct. You could have used plain comparison, since $sin(k+5)$ is bounded and $(3/2)^{-k}$ goes to zero faster than, say, $k^{-4}$.
    $endgroup$
    – GReyes
    Jan 29 at 7:35










  • $begingroup$
    @GReyes what exactly our comparison function would look like?Probably ${k^2over k^4}={1over k^2}$?
    $endgroup$
    – Turan Nasibli
    Jan 29 at 7:37












  • $begingroup$
    I think your sum does converge.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 29 at 7:49










  • $begingroup$
    @Dr.SonnhardGraubner it does,for sure.
    $endgroup$
    – Turan Nasibli
    Jan 29 at 7:51










  • $begingroup$
    @TuranNəsibli Yes, you are right. For large enough $k$, $(3/2)^{-k}le k^{-4}$ and $sin(k+5)le 1$, therefore your $|a_k|le 1/k^2$.
    $endgroup$
    – GReyes
    Jan 29 at 20:04














  • 2




    $begingroup$
    Yes, this is correct. You could have used plain comparison, since $sin(k+5)$ is bounded and $(3/2)^{-k}$ goes to zero faster than, say, $k^{-4}$.
    $endgroup$
    – GReyes
    Jan 29 at 7:35










  • $begingroup$
    @GReyes what exactly our comparison function would look like?Probably ${k^2over k^4}={1over k^2}$?
    $endgroup$
    – Turan Nasibli
    Jan 29 at 7:37












  • $begingroup$
    I think your sum does converge.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 29 at 7:49










  • $begingroup$
    @Dr.SonnhardGraubner it does,for sure.
    $endgroup$
    – Turan Nasibli
    Jan 29 at 7:51










  • $begingroup$
    @TuranNəsibli Yes, you are right. For large enough $k$, $(3/2)^{-k}le k^{-4}$ and $sin(k+5)le 1$, therefore your $|a_k|le 1/k^2$.
    $endgroup$
    – GReyes
    Jan 29 at 20:04








2




2




$begingroup$
Yes, this is correct. You could have used plain comparison, since $sin(k+5)$ is bounded and $(3/2)^{-k}$ goes to zero faster than, say, $k^{-4}$.
$endgroup$
– GReyes
Jan 29 at 7:35




$begingroup$
Yes, this is correct. You could have used plain comparison, since $sin(k+5)$ is bounded and $(3/2)^{-k}$ goes to zero faster than, say, $k^{-4}$.
$endgroup$
– GReyes
Jan 29 at 7:35












$begingroup$
@GReyes what exactly our comparison function would look like?Probably ${k^2over k^4}={1over k^2}$?
$endgroup$
– Turan Nasibli
Jan 29 at 7:37






$begingroup$
@GReyes what exactly our comparison function would look like?Probably ${k^2over k^4}={1over k^2}$?
$endgroup$
– Turan Nasibli
Jan 29 at 7:37














$begingroup$
I think your sum does converge.
$endgroup$
– Dr. Sonnhard Graubner
Jan 29 at 7:49




$begingroup$
I think your sum does converge.
$endgroup$
– Dr. Sonnhard Graubner
Jan 29 at 7:49












$begingroup$
@Dr.SonnhardGraubner it does,for sure.
$endgroup$
– Turan Nasibli
Jan 29 at 7:51




$begingroup$
@Dr.SonnhardGraubner it does,for sure.
$endgroup$
– Turan Nasibli
Jan 29 at 7:51












$begingroup$
@TuranNəsibli Yes, you are right. For large enough $k$, $(3/2)^{-k}le k^{-4}$ and $sin(k+5)le 1$, therefore your $|a_k|le 1/k^2$.
$endgroup$
– GReyes
Jan 29 at 20:04




$begingroup$
@TuranNəsibli Yes, you are right. For large enough $k$, $(3/2)^{-k}le k^{-4}$ and $sin(k+5)le 1$, therefore your $|a_k|le 1/k^2$.
$endgroup$
– GReyes
Jan 29 at 20:04










0






active

oldest

votes












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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3091881%2fconvergence-of-a-k-sum-k-5-infty-1k3-over2-kk25-sink5%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3091881%2fconvergence-of-a-k-sum-k-5-infty-1k3-over2-kk25-sink5%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

MongoDB - Not Authorized To Execute Command

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
Read more

Npm cannot find a required file even through it is in the searched directory

Image
up vote 0 down vote favorite I got that error: users-Air:pdfuploader user$ npm run start client@0.1.0 start /Users/user/Documents/pdfuploader react-scripts start Could not find a required file. Name: index.html Searched in: /Users/user/Documents/pdfuploader/public npm ERR! code ELIFECYCLE npm ERR! errno 1 npm ERR! client@0.1.0 start: `react-scripts start` npm ERR! Exit status 1 npm ERR! npm ERR! Failed at the client@0.1.0 start script. npm ERR! This is probably not a problem with npm. There is likely additional logging output above. npm ERR! A complete log of this run can be found in: npm ERR! /Users/user/.npm/_logs/2018-11-19T10_19_14_555Z-debug.log users-Air:pdfuploader user$ And the file is in /Users/user/Documents/pdfuploader/public as you can see in my github repository: https://github.c...
Read more

How to fix TextFormField cause rebuild widget in Flutter

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
Read more