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







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$endgroup$








  • 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






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share|cite|improve this question













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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











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1 Answer
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active

oldest

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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


















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