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I am trying to find the closed form of the following related sums:

$$(i)\quad\quad S_1(n)= \sum_{m=-\infty}^{m=\infty} |n-m| e^{-p(|n-m|+|m|)} $$

$$ (ii)\quad\quad S_2(n)= \sum_{m=-\infty}^{m=\infty} m(|n-m|+\gamma) e^{-p(|n-m|+|m|)} $$

$$ (ii) \quad \quad S_3(n)= \sum_{m=-\infty}^{m=\infty} (n-m)^2 e^{-p(|n-m|+|m|)} $$

where $\gamma,p\in \mathbb{R}^+$.

Any tips are appreciated.

Nick P
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1 Answers1

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For (i) let's define $f(x) = xe^{-px}$ for $x\ge 0.$ The sum takes the form

$$S(n)=\sum_{m\in \mathbb {Z}}f(|n-m|)e^{-p|m|}.$$

Now $e^{-p|m|}\in l^1(\mathbb {Z})..$ Because $f$ is bounded on $[0,\infty),$ and $f(x) \to 0$ at $\infty,$ we are set up to use the dominated convervence theorem:

$$\lim_{n\to \infty} S(n) = \sum_{m\in \mathbb {Z}}[\lim_{n\to \infty}f(|n-m|)]e^{-p|m|} =\sum_{m\in \mathbb {Z}}0=0$$

Try this idea on the other ones.

zhw.
  • 105,693
  • The question is unclear, and I have edited, but I want $S(n)$ for all $n$, not just for $n\to \infty$. Apologies for the confusion. – Nick P May 26 '15 at 00:26