Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Here is "almost an integer" result:

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

?

share|improve this question
    
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
add comment

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 $163$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.