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I've been solving a Renewal theory problem and I end up with this function

$m(t)=e^{-4t}\sum_{k=1}^{\infty}\sum_{i=2k}^{\infty}\frac{(4t)^i}{i!}$. How do I solve or reduce the double summation? Is it justifiable to swap them? why/why not? It seems like if I want to reduce the inside expression and put it in terms of $f\rightarrow f(e^{4t},t)$ is giving me trouble, and I am ending up with a recursion that I don't know how to handle. I think the answer might be found if I use a Generating funtion, but I'm not sure.

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A given term $(4t)^i/i!$ occurs for $k=1,\dots, i/2,$ so you have the sum break up into two pieces (for odd and even $i$) The even part looks like $\sum_{i=1}^\infty (4t)^i/(2(i-1)!),$ the odd part is very slightly more complicated, but both are easy to evaluate.

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  • $\begingroup$ But what I want to know is how to get rid of the sums, and leave the equation in a reduced form, and/or if that is even possible. $\endgroup$ – Arturo Dec 1 '13 at 21:39

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