EDIT, Will Jagy, December 8, 2010: to anyone considering working on this, please first see http://mathoverflow.tqft.net/discussion/817/could-a-few-moderators-please-remove-one-of-my-questions/#Item_9
which gives the story behind this peculiar sum. Note that the OP is no longer interested in the results, as they arose from one kind of error and cannot be applied because of a different sort of misunderstanding. The double sum version below was provided recently by Harald Hanche-Olsen.
ORIGINAL. I'm curious one of you is able to find the exact evaluation of
the following series:
$$\begin{aligned} S &= 1/(2\times3) +1/(5\times6) + 1/(7\times8) + 1/(10\times11) + \cdots
\\\\&= \sum_{n=1}^\infty\sum_{k=1}^{n}\frac1{(n^2+2k-1)(n^2+2k)} \end{aligned}$$
I'm not exactly sure on how to state the 'general term' of the series. Perhaps I can illustrate it with an example:
$ 1/(1\times2) + 1/(3\times4) + 1/(5\times6) + 1/(7\times8) + \ldots + 1/((2n – 1) \times 2n) + \ldots = \log(2)$.
Now, to answer Nate Eldredge: let $a_0=2$ and $a_{k+1}=a_{k} + 1 $ unless $ a_{k} + 1$ is a square, in which case let $a_{k + 1} = a_{k} + 2$. Now, multiply $a_{k}$ with $a_{k+1}$. That's a term. Let me show the first few terms:
$ S = 1/(2\times3)$ [now skip 4] $ + 1/(5\times6) + 1/(7\times8)$ [now skip 9] $ + 1/(10\times11) + 1/(12\times13) + 1/(14\times15)$ [now skip 16] $ + 1/(17\times18) + \ldots $
So all the squares (1,4,9,16,25, etc) are 'skipped' in the terms.
I hope this clarifies it a bit…
Thanks a lot in advance,
Max Muller
PS: If someone has any ideas as to how the general term of this series can be written in a more concise manner, please let me know! For the Meta-users, see also the relevant discussion on this question.
Best Answer
I thought I'd add in my favorite way of computing sums of the form $$\sum_{n=0}^{\infty} \frac{p(n)}{q(n)}$$ for $p$ and $q$ polynomials.
First, expand in partial fractions: $$\frac{p(n)}{q(n)} = \sum \frac{a_i}{n+s_i} + \mbox{terms for repeated roots}.$$
Since the sum is to converge, we must have $p(n)/q(n) =O(n^{-2})$, so $\sum a_i=0$. So rewrite this as $$\frac{p(n)}{q(n)} = \sum a_i \left( \frac{1}{n+s_i} - \frac{1}{n} \right) + \mbox{terms for repeated roots}.$$
We are now reduced to evaluating sums of the form $$\sum \left( \frac{1}{n} - \frac{1}{n+s} \right)$$ and $$\sum \frac{1}{(n+s)^k}$$
We have the identity $$\sum \left( \frac{1}{n} - \frac{1}{n+s}\right) = \frac{\Gamma'(s)}{\Gamma(s)} + \gamma + \frac{1}{s}$$ where $\Gamma$ is the Gamma function and $\gamma$ is the Euler–Mascheroni constant. See, for example, this website. Taking repeated derivatives of this gives a formula for $\sum 1/(n+s)^k$.
So, if you accept derivatives of the $\Gamma$ function at algebraic numbers, such sums can always be evaluated. In general, I don't know a better method. However, in many cases, one can use familiar $\Gamma$ function identities to do better. Here are three tricks, all of which come up in our setting:
If we ever have to deal with $\sum \left( \frac{1}{n+s} - \frac{1}{n+k+s} \right)$, then the sum telescopes. We can see this on the $\Gamma$ function side: $\Gamma(s+k) = (s+k-1)\cdots (s+1) (s) \Gamma(s)$ and taking logarithimic derivatives gives a relation between $\Gamma(s+k)/\Gamma'(s+k)$ and $\Gamma'(s)/\Gamma(s)$.
If we ever have to deal with $\sum \left( \frac{1}{n-s} - \frac{1}{n+s} \right)$, recall that $$\Gamma(s) \Gamma(-s) = \frac{\pi}{s \sin (\pi s)}.$$ Taking logarithimic derivatives of this will give you a formula for $\Gamma'(s)/\Gamma(s) - \Gamma'(-s)/\Gamma(-s)$. Repeated derivatives will deal with $\sum \left( \frac{1}{(n-s)^k} - \frac{1}{(n+s)^k} \right)$. It is important to note that all of these formulas work for $s$ complex; for example, in the current example we need to deal with $s=i/2$.
Remember that $\zeta(2k)$ is already known, and is easy to look up that than the Taylor series of $\Gamma$ around $0$.