Bound for summation

  • Thread starter bruno67
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  • #1
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Main Question or Discussion Point

I am looking for a bound for the following expression

[tex]S=\sum_{n=1}^N n^k e^{-an}[/tex]
where a>0 and k=1, 2, 3, or 4, apart from the obvious one:

[tex]S\le \frac{n+1}{2} \sum_{n=1}^N e^{-an} = \frac{n+1}{2}
\frac{1-e^{-Na}}{e^a-1}[/tex]
 
Last edited:

Answers and Replies

  • #2
32
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I got it. S(k) is bound by the integral

[tex]S\le \int_1^{N+1} x^k e^{-ax} dx[/tex]
 

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