Thread View: alt.uu.math.misc
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Started by Enrico Talinucci
Fri, 03 Dec 1999 00:00
Upper bound for \sum{k=n}^\infty (x^k / k!)
Author: Enrico Talinucci
Date: Fri, 03 Dec 1999 00:00
Date: Fri, 03 Dec 1999 00:00
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1451 bytes
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
Re: Upper bound for \sum{k=n}^\infty (x^k / k!)
Author: Pierre Asselin
Date: Fri, 03 Dec 1999 00:00
Date: Fri, 03 Dec 1999 00:00
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"Enrico Talinucci" <talinucci@tin.it> writes:
>I need to upper bound the following sum:
>S(n) = \sum_{k=n}^\infty (x^k/k!)
Hint: for k>n, x^k/k! is less than (x^n/n!) * (x/n)^(k-n) .
--
--Pierre Asselin, Westminster, Colorado
lpa@netcom.com
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