Convergence of sum using ratio test












1












$begingroup$


Determine the values of x ∈ R for which the following series converge:



$sum_1^infty frac{x^nn^n}{n!}$



My attempt: I used the ratio test to obtain $|x|< frac{1}{e}$ as the interval where it convergences for sure, but I am unable to determine whether the series convergences at $x=frac{1}{e}$ and $x=frac{-1}{e}$ (when the limit of the ratio is $1$ and the test is inconclusive).






sequences-and-series absolute-convergence







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Try Stirling's formula.
    $endgroup$
    – Lord Shark the Unknown
    Dec 11 '18 at 7:36
















1












$begingroup$


Determine the values of x ∈ R for which the following series converge:



$sum_1^infty frac{x^nn^n}{n!}$



My attempt: I used the ratio test to obtain $|x|< frac{1}{e}$ as the interval where it convergences for sure, but I am unable to determine whether the series convergences at $x=frac{1}{e}$ and $x=frac{-1}{e}$ (when the limit of the ratio is $1$ and the test is inconclusive).






sequences-and-series absolute-convergence







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Try Stirling's formula.
    $endgroup$
    – Lord Shark the Unknown
    Dec 11 '18 at 7:36














1












1








1


1



$begingroup$


Determine the values of x ∈ R for which the following series converge:



$sum_1^infty frac{x^nn^n}{n!}$



My attempt: I used the ratio test to obtain $|x|< frac{1}{e}$ as the interval where it convergences for sure, but I am unable to determine whether the series convergences at $x=frac{1}{e}$ and $x=frac{-1}{e}$ (when the limit of the ratio is $1$ and the test is inconclusive).






sequences-and-series absolute-convergence







share|cite|improve this question











$endgroup$




Determine the values of x ∈ R for which the following series converge:



$sum_1^infty frac{x^nn^n}{n!}$



My attempt: I used the ratio test to obtain $|x|< frac{1}{e}$ as the interval where it convergences for sure, but I am unable to determine whether the series convergences at $x=frac{1}{e}$ and $x=frac{-1}{e}$ (when the limit of the ratio is $1$ and the test is inconclusive).






sequences-and-series absolute-convergence




sequences-and-series absolute-convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 7:40









nafhgood

1,805422




1,805422










asked Dec 11 '18 at 7:33









childishsadbinochildishsadbino

1148




1148








  • 1




    $begingroup$
    Try Stirling's formula.
    $endgroup$
    – Lord Shark the Unknown
    Dec 11 '18 at 7:36














  • 1




    $begingroup$
    Try Stirling's formula.
    $endgroup$
    – Lord Shark the Unknown
    Dec 11 '18 at 7:36








1




1




$begingroup$
Try Stirling's formula.
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 7:36




$begingroup$
Try Stirling's formula.
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 7:36










1 Answer
1






active

oldest

votes


















1












$begingroup$

For the cases $x=pm frac 1 e$ we can use Stirling approximation



$$n! sim sqrt{2 pi n}left(frac{n}{e}right)^n$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So, that gives me that at x=1/e, series diverges by p-test and at x=-1/e, it converges by alternating series test. Is this correct?
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:46










  • $begingroup$
    @childishsadbino That's absolutely correct indeed!
    $endgroup$
    – gimusi
    Dec 11 '18 at 7:47











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%2f3035012%2fconvergence-of-sum-using-ratio-test%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









1












$begingroup$

For the cases $x=pm frac 1 e$ we can use Stirling approximation



$$n! sim sqrt{2 pi n}left(frac{n}{e}right)^n$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So, that gives me that at x=1/e, series diverges by p-test and at x=-1/e, it converges by alternating series test. Is this correct?
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:46










  • $begingroup$
    @childishsadbino That's absolutely correct indeed!
    $endgroup$
    – gimusi
    Dec 11 '18 at 7:47
















1












$begingroup$

For the cases $x=pm frac 1 e$ we can use Stirling approximation



$$n! sim sqrt{2 pi n}left(frac{n}{e}right)^n$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So, that gives me that at x=1/e, series diverges by p-test and at x=-1/e, it converges by alternating series test. Is this correct?
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:46










  • $begingroup$
    @childishsadbino That's absolutely correct indeed!
    $endgroup$
    – gimusi
    Dec 11 '18 at 7:47














1












1








1





$begingroup$

For the cases $x=pm frac 1 e$ we can use Stirling approximation



$$n! sim sqrt{2 pi n}left(frac{n}{e}right)^n$$






share|cite|improve this answer









$endgroup$



For the cases $x=pm frac 1 e$ we can use Stirling approximation



$$n! sim sqrt{2 pi n}left(frac{n}{e}right)^n$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 11 '18 at 7:36









gimusigimusi

92.8k84494




92.8k84494












  • $begingroup$
    So, that gives me that at x=1/e, series diverges by p-test and at x=-1/e, it converges by alternating series test. Is this correct?
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:46










  • $begingroup$
    @childishsadbino That's absolutely correct indeed!
    $endgroup$
    – gimusi
    Dec 11 '18 at 7:47


















  • $begingroup$
    So, that gives me that at x=1/e, series diverges by p-test and at x=-1/e, it converges by alternating series test. Is this correct?
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:46










  • $begingroup$
    @childishsadbino That's absolutely correct indeed!
    $endgroup$
    – gimusi
    Dec 11 '18 at 7:47
















$begingroup$
So, that gives me that at x=1/e, series diverges by p-test and at x=-1/e, it converges by alternating series test. Is this correct?
$endgroup$
– childishsadbino
Dec 11 '18 at 7:46




$begingroup$
So, that gives me that at x=1/e, series diverges by p-test and at x=-1/e, it converges by alternating series test. Is this correct?
$endgroup$
– childishsadbino
Dec 11 '18 at 7:46












$begingroup$
@childishsadbino That's absolutely correct indeed!
$endgroup$
– gimusi
Dec 11 '18 at 7:47




$begingroup$
@childishsadbino That's absolutely correct indeed!
$endgroup$
– gimusi
Dec 11 '18 at 7:47


















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%2f3035012%2fconvergence-of-sum-using-ratio-test%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

Aardman Animations

Image
Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com
Read more

Are they similar matrix

Image
0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19
Read more

“minimization” problem in Euclidean space related to orthonormal basis

Image
3 $begingroup$ I have stumbled upon an exercice for second year undegraduate student majoring in economics which I find quite demanding. I have an idea for the solution, but it seems awfully complicated, and I am wondering if there was a simpler way to solve it. This goes as follows: Let $(E, langle, rangle)$ be a euclidean space of dimension $n geq 2$ . 1) For any $x in E$ , show that $||x|| = sqrt{n}$ if and only if there exists $(e_1, dots, e_n)$ an orthonormal basis of $E$ such that $x =e_1+ ldots + e_n$ . 2) For any $(x,y) in E$ , show that $||x|| = sqrt{n}$ , $||y|| = sqrt{frac{n(n+1)(2n+1)}{6}}$ and $langle x,y rangle = frac{n(n+1)}{2}$ if and only if there eixsts $(e_1, ldots, e_n)$ an orthonormal basis of $E$ such that $x = e_1 + ldots + e_n$ and $y = e_1 + 2e_2 + ldots + ne_n$ The first q
Read more