4
$\begingroup$

Here is "almost an integer" result:

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

?

  • 0
    I hope I did not err in changing your expression "(1/(exp(Pi*sqrt(163))^k))" to $e^{-k\pi\sqrt{163}}$.2012-06-24
  • 0
    @Jyrki I would vote for these comments as an answer.2012-06-24

1 Answers 1

26

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 $163$.

  • 0
    Well, there is, of course, a chance that the series is very interesting for some other reason, but do you know of one?2013-05-20