Bound for S: Sum of n^k e^(-an)

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The discussion focuses on deriving a bound for the expression S=\sum_{n=1}^N n^k e^{-an} where a>0 and k is an integer from 1 to 4. The initial bound presented is S ≤ \frac{n+1}{2} \sum_{n=1}^N e^{-an}, which simplifies to S ≤ \frac{n+1}{2} \frac{1-e^{-Na}}{e^a-1}. Additionally, it is established that S(k) can be further bounded by the integral S ≤ \int_1^{N+1} x^k e^{-ax} dx, providing a more refined approach to estimating the sum.

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bruno67
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I am looking for a bound for the following expression

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

S\le \frac{n+1}{2} \sum_{n=1}^N e^{-an} = \frac{n+1}{2}<br /> \frac{1-e^{-Na}}{e^a-1}
 
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I got it. S(k) is bound by the integral

S\le \int_1^{N+1} x^k e^{-ax} dx
 

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