Re: Unknown Sum of Series
- To: mathgroup at smc.vnet.net
- Subject: [mg63680] Re: Unknown Sum of Series
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Mon, 9 Jan 2006 04:50:10 -0500 (EST)
- Organization: The University of Western Australia
- References: <dpb3vu$9tg$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <dpb3vu$9tg$1 at smc.vnet.net>,
"Klaus G." <Karl_boehme_9 at msn.com> wrote:
> Mathematica 5.0 is not able to compute the symbolic sum:
>
> Sum[(-1)^(1 + n)*(E - ( 1 + (1/n))^n ), {n, 1, Infinity}]
>
> However, Nsum[...] results in 0.4456224031968407..
>
> I tried http://oldweb.cecm.sfu.ca/projects/ISC/ to find hidden
> constants in that number like Pi or E, but without success.
>
> Any idea?
Since
(1 + (1/n))^n == Exp[n Log[1 + 1/n]]
then using the generating function for Stirling Numbers of the first kind
http://functions.wolfram.com/04.14.11.0005.01
I can find a formal representation for this sum as
(1/2) Log[2] E +
Sum[StirlingS1[p + l, l]/(p + l)! (2^(1 - p) - 1) Zeta[p],
{l, 0, Infinity}, {p, 2, Infinity}]
However, the convergence of this double sum is very slow and is probably
not useful to you.
Cheers,
Paul
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Paul Abbott Phone: 61 8 6488 2734
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