<s>[INST] Evaluation of the following series... $S = 1/(2\times3) + 1/(5\times6) + 1/(7\times8) + 1/(10\times11) + ...  $ 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.
 [/INST] As you are interested in $ \zeta(3) $ you might prefer this variant of your construction. This is also a small part of what Pietro Majer would have done in the direction indicated by Charles Matthews.
Define $ f(x)$ for $ | x | \leq 1 $ by
$$  
 f(x)   =  \frac{x^4}{2 \cdot 3 \cdot 4} + \frac{x^7}{5 \cdot 6 \cdot 7}   +  \frac{x^{11}}{9 \cdot 10 \cdot 11}  +  \frac{x^{14}}{12 \cdot 13 \cdot 14}  +\cdots + \frac{x^{30}}{28 \cdot 29 \cdot 30} +  \frac{x^{33}}{31 \cdot 32 \cdot 33} + \cdots
    $$
Then I took the third derivative, power series has all coefficients 1 and simplifies as 
$$ f'''(x) = (x + x^4 ) + ( x^8 + x^{11} + \cdots + x^{23}) + ( x^{27} + x^{30} + \cdots) + \cdots $$
$$ f'''(x) = \sum_{n=1}^\infty \; \; x^{n^3}   \left( \frac{1 - x^{3 n^2 + 3 n}}{1 - x^3}   \right)  $$ 
or
$$ \left( \frac{1 }{1 - x^3}   \right) \cdot \left( \sum_{n=1}^\infty \; \; x^{n^3} - x^{n^3 + 3 n^2 + 3 n} \right) $$
$$ \left( \frac{1 }{1 - x^3}   \right) \cdot \left( \sum_{n=1}^\infty \; \; x^{n^3} - \left( \frac{1}{x}   \right)  \sum_{n=1}^\infty \; \; x^{(n + 1)^3} \right) $$
$$ \left( \frac{1 }{1 - x^3}   \right) \cdot \left( \sum_{n=1}^\infty \; \; x^{n^3} - \left( \frac{1}{x}   \right)  \sum_{m=2}^\infty \; \; x^{m^3} \right) $$
$$ \left( \frac{1 }{1 - x^3}   \right) \cdot \left( 1 + \sum_{n=1}^\infty \; \; x^{n^3} - \left( \frac{1}{x}   \right)  \sum_{m=1}^\infty \; \; x^{m^3} \right) $$
$$ \left( \frac{1 }{1 - x^3}   \right) \cdot \left( 1  - \left( \frac{1-x}{x}   \right)  \sum_{m=1}^\infty \; \; x^{m^3} \right) $$
$$ f'''(x) = \left( \frac{1 }{1 - x^3}   \right) - \left( \frac{1}{x ( 1 + x + x^2)}   \right)  \sum_{m=1}^\infty \; \; x^{m^3}  $$
And so on more or less forever, with no explicit value for $f(1)$ likely. 
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