Upper bound for \sum{k=n}^\infty (x^k / k!)

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#27

From: Enrico Talinucci
Date: Fri, 03 Dec 1999 00:00

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Hi everybody!

I need to upper bound the following sum:

S(n) = \sum_{k=n}^\infty (x^k/k!)

The generalized mean value theorem (Lagrange remainder) says

S(n) < S'(n) = exp(x) (x^{n+1}/(n+1)!)

I'm looking for something better. S(n) is strictly decreasing with respect
to n,
but the the Lagrange bound S'(n) is not. More precisely, S'(n) grows for
n < x-1, has a maximum and then decreases.

Does anybody know a deceasing bound for S(n), if possible stronger than
Lagrange's
for every n?

Any help will be really appreciated. Thanks in advance.

Enrico Talinucci

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From: "Enrico Talinucci" <talinucci@tin.it>
Subject: Upper bound for \sum{k=n}^\infty (x^k / k!)
Date: 1999/12/03
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