Re: Does this series converge?

In article <250120060134477958%bruck@xxxxxxxxxxxxxxxxx>,
Ronald Bruck <bruck@xxxxxxxxxxxxxxxxx> wrote:
>In article <r9HBf.2276$v81.1009@xxxxxxxx>, Stephen J. Herschkorn
><sjherschko@xxxxxxxxxxxx> wrote:
>
>> Does the series, sum(n=0..infty, (n+1)^n exp(-n) / n!) converge? How
>> can we tell?
>
>No, it doesn't. Use Stirling's formula and the fact that (1+1/n)^n
>converges to e to show that
>
> sqrt(n) * your term --> e/sqrt(2 pi) > 1,
>
>and therefore your n-th term is > 1/sqrt(n).

In fact, if we use more terms in Stirling's asymptotic series, we get
that the n^{th} term is asymptotically

e 1 7 145
--------- ------- ( 1 - --- + ------ + ... )
sqrt(2pi) sqrt(n) 12n 288n^2

This allows us to say that the alternating series does converge.
Just being asymptotic to 1/sqrt(n) is not enough; for example

oo
--- (-1)^n 1
> ------- + -
--- sqrt(n) n
n=1

is an alternating series, the absolute value of whose terms are
asymptotic to 1/sqrt(n), which diverges.

Rob Johnson <rob@xxxxxxxxxxxxxx>
take out the trash before replying
.

Relevant Pages

  • Re: Where do I begin?
    ... Wouldn't the squeeze theorem work? ... are alternating series, uniformly decreasing and the n'th term goes to ... Prev by Date: ...
    (sci.math)
  • Re: Where do I begin?
    ... > Wouldn't the squeeze theorem work? ... > are alternating series, uniformly decreasing and the n'th term goes to ... Prev by Date: ...
    (sci.math)