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Here is "almost an integer" result:

$$\sum^{\infty}_{k=0}\left(\frac{1}{\exp(\pi\sqrt{163})}\right)^{k}\left(\frac{120}{8k+1}-\frac{60}{8k+4}-\frac{30}{8k+5}-\frac{30}{8k+6}\right) = 94.000000000000000014789449792044364408558923807659819...$$


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I hope I did not err in changing your expression "(1/(exp(Pi*sqrt(163))^k))" to $e^{-k\pi\sqrt{163}}$. –  MJD Jun 24 '12 at 17:25
@Jyrki I would vote for these comments as an answer. –  alex.jordan Jun 24 '12 at 17:44
@alex, I moved the comments to an answer. –  Jyrki Lahtonen Jun 24 '12 at 17:54

1 Answer 1

There is nothing magical about this sum. Remember that $e^{-\pi\sqrt{163}}\approx 4\cdot10^{-18}$ is a small positive number. When you substitute $k=0$, you get the main term $=94$. The other terms are all tiny.

If you don't believe this, try the following. Compute the same sums with $164, 165,\ldots$ instead of $163$.

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Well, there is, of course, a chance that the series is very interesting for some other reason, but do you know of one? –  Jyrki Lahtonen May 20 '13 at 4:14

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