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I would like to know if there a closed form solution for the sum:

$ S_n(t) = \sum_{k=0}^{n} \cos( t \sqrt{k} ) $

There is obviously an easy answer when the sum is replaced by an integral so this question is really asking for the exact form that $S_n(t)$ takes.

Any hints or references on how to approach this problem would be welcome.

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    Terminological nitpick - one does not speak of "solution for the sum," rather, of "evaluation for the sum."2011-05-07

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$S_{n}=\sum^{n}_{k=0}\cos(t\sqrt{k})=\sum^{n}_{k=0}\frac{e^{it\sqrt{k}}+e^{-it\sqrt{k}}}{2}$. Any closed form of this must somehow evaluate $\sum^{n}_{k=0}e^{it\sqrt{k}}$. Consider $e^{x}=\sum^{\infty}_{i=0}x^{i}/i!$, $e^{it\sqrt{k}}$ would be $\sum^{\infty}_{j=0}(it\sqrt{k})^{j}/j!$. Hence the series would be:

$+_{n}(t)=\sum^{\infty}_{k=0}\sum^{\infty}_{j=0}(it\sqrt{k})^{j}/j!$.

This can be decomposed into even and odd $j$ terms. Since $j$ evaluate from $0$ we name the even one to be $S_{0}$ and the odd one $S_{1}$. The even terms $j=2s$ sums up to:

$+_{0}(t)=\sum^{\infty}_{k=0}\sum^{\infty}_{s=0}-(t^{2s}k^{s})/(2s)!$.

And the odd one $j=2s+1$ sums up to:

$+_{1}(t)=-i\sum^{\infty}_{k=0}\sum^{\infty}_{s=0}(t\sqrt{k})^{2s+1}/(2s+1)!$

Now, what is the closed form of the original series? It is the same as $+_{0}(t)$. Switch $k$ and $s$ we should have $\sum^{\infty}_{s=0}\sum^{\infty}_{k=0}-[(t^{2s})/(2s)!]k^{s}$

I think there is NO such closed form to evaluate this double series.

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    Quite a long-winded answer, but finally you come to the correct answer to the question: "No".2011-05-07