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.
 A: NB: It seems that after this answer was written, the asker "made precise" what he wanted, and I think this has been rendered more or less irrelevant...
NB2: Maybe the people voting down this can explain their votes?
Assuming you want to sum the terms of $\displaystyle\sum_{n\geq1}\frac1{n(n+1)}$ which are not of the form $\displaystyle\frac{1}{n^2(n^2+1)}$ or $\displaystyle\frac{1}{(n^2-1)n^2}$, you can do it by computing the first sum (by the telescoping trick) and then substracting the sum of the terms you want to exclude. To compute the sums you need to substract, say, $\displaystyle\sum_{n\geq2}\frac{1}{(n^2-1)n^2}$, you can compute it as the sum of residues of $\frac{\cot z}{z^4-z^2}$ at $2$, $3$, $4$, etc, using a bit of complex analysis (as in the 8th solution of the Basel problem presented by R. Chapman here)
A: As already mentioned by Mariano, the sum in question is a simple logarithmic sum minus
$$
\sum_{n>1}\biggl(\frac1{n^2(n^2-1)}+\frac1{n^2(n^2+1)}\biggr)
=2\sum_{n>1}\frac1{n^4-1}
=\frac74-\frac\pi2\coth\pi.
$$
The closed form evaluation is mentioned in Michel Waldschmidt's
"Open Diophantine Problems" (Moscow Math. Journal 4:1 (2004), 245-305, 312) together with "which is a transcendental number since $\pi$ and $e^\pi$ are algebraically independent over $\mathbb Q$ (Yu.V. Nesterenko)."
EDIT. I was confused by too many answers and edits. On using
$$
\frac1{n(n+1)}=\frac1n-\frac1{n+1},
$$
one can write the wanted sum as
$$
\sum_{n=1}^\infty\frac{(-1)^{n+[\sqrt n]-1}}n+\sum_{k=1}^\infty\frac1{k^2}
=\sum_{n=1}^\infty\frac{(-1)^{n+[\sqrt n]-1}}n+\frac{\pi^2}6.
$$
I can hardly imagine that the remaining single sum has a closed form... But this leaves a nice problem in 1st year analysis: Show that the series
$$
\sum_{n=1}^\infty\frac{(-1)^{n+[\sqrt n]}}n
$$
converges.
A: rmk 1. After some manipulation of the series, if
$f(x):=1 - \frac{1-x}{1+x}\sum_{k\in\mathbb Z} x^{k^2}$
Then 
$S:=\frac{1}{2}\int_0^1\frac{f(x)}{x} dx.$
So this confirms Charles Matthews' hint: 
$\theta:=\sum_{k\in\mathbb Z} x^{k^2}$
Is the theta series, indeed, linked to the Jacobi theta function. (But I don't know how to proceed then.)
rmk 2. Maybe writing Max Muller's series as a sum over double indices (k,n) 
$S:=\sum_{1\le k\le n} \frac{1}{(n^2+2k-1)(n^2+2k)}$
one can find a smart partition of the set of indices that allows an iterate summation. (very few chances of success). For instance changing the order of summation, i.e.
$S:=\sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{(n^2+2k-1)(n^2+2k)}$
The inner sum can be treated via partial fraction decomposition and summed as illustrated by David Speyer; I found (with $\Psi:=\Gamma'/\Gamma$)
$\sum_{k=1}^\infty\ \left( \mathrm{Im}\frac{\Psi(k+i\sqrt{2k-1})}{\sqrt{2k-1}}-\mathrm{Im}\frac{\Psi(k+i\sqrt{2k})}{\sqrt{2k}}\right)=0.2666107..$
A: 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. 
A: Anyway, I think the point is that with a(k) a sequence of the type implicated here, namely blocks of +1 and -1 alternating, jumping and flipping over at k any square, the generating function of the a(k) is composed of blocks of geometric series. If the variable is t, each finite geometric series can be summed with denominator 1 + t. At this stage I think Euler could handle the question? Rational function times a theta-series, I believe. This needs to be integrated formally, and the limit as t -> 1 from below taken. This would be the "classical" approach.
I apologise if this is nonsense. 
A: 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 $0$. 
A: Incorrect answer deleted.
A: S=1/(2×3)+1/(5×6)+1/(7×8)+1/(10×11)+...=1/20(20-2(pi)Sqrt(5-2Sqrt(5))
-2Sqrt(5)ArcCoth(Sqrt(5))-5Log(5))=0.2617882198...(approx.)
