Can this series be put in a generalized form? I asked a similar question here But this one seems to not work out so nicely...
I started looking at the series,
$$
S = \frac{1}{2}+\frac{2}{3}-\frac{3}{4}-\frac{4}{5}+\frac{5}{6}+\frac{6}{7}-\frac{7}{8}-\frac{8}{9}+\cdots
$$
Which is equivilent to
$$
S = \sum_{k=0}^{\infty}[\frac{4k+1}{4k+2}+\frac{4k+2}{4k+3}-\frac{4k+3}{4k+4}-\frac{4k+4}{4k+5}]
$$
I found $S = \frac{\pi}{4}-\frac{\ln(4)}{4}-1$
Then i put the series in a different form such that
$$
S_n = \sum_{k=0}^\infty[\frac{4k+n}{4k+n+1}+\frac{4k+n+1}{4k+n+2}-\frac{4k+n+2}{4k+n+3}-\frac{4k+n+3}{4k+n+4}]
$$
$$
S_n = -\sum_{k=0}^{\infty}\frac{2(2n^2 + 16nk+10n+32k^2+40k+11)}{(4k+n+1)(4k+n+2)(4k+n+3)(4k+n+4)}
$$
Going back to the question i had previously asked i learned that $\psi(1+x)=\sum_{k=1}^\infty \frac{x}{k(k+x)}$ which does not seem to apply here. The series yields some interesting results
$$
S_1 = \frac{\pi}{4}-\frac{\ln(4)}{4}-1
$$
$$
S_2 = \frac{\pi}{4}+\frac{\ln(4)}{4}-\frac{3}{2}
$$
$$
S_3=\frac{\ln(4)}{4}+\frac{1}{6}-\frac{\pi}{4}
$$
$$
S_4 = \frac{11}{12}-\frac{\pi}{4} - \frac{\ln(4)}{4}
$$
$$
S_5 = \frac{\pi}{4}-\frac{\ln(2)}{2}-\frac{37}{60}
$$
It seems like $\frac{pi}{4}$ occurs for every $n$ value, I would like to know if there is a generalized form of this series like the result i found here and how i would go about identifying this form (if one exists).

This is nowhere near a complete answer, so I've marked it community wiki. 

$$\frac{n-1}{n} + \frac{n}{n+1} = \frac{n^2 -1}{n(n+1)} + \frac{n^2}{n(n+1)} = \frac{2n^2-1}{n^2+n}$$
This means every pair of consecutive terms in your original series that have the same sign sum to $$\pm \frac{2n^2 -1}{n^2+n}$$ for some $n \in \mathbb{Z}$. This gives an alternate way to write your sum: 
$$S = \left(\frac{1}{2} + \frac{2}{3}\right) - \left(\frac{3}{4} + \frac{4}{5}\right) + \left(\frac{5}{6} + \frac{6}{7}\right) - \cdots $$
$$S = \frac{2(2^2)-1}{2^2+2} - \frac{2(4^2)-1}{4^2+4}+\frac{2(6^2)-1}{6^2+6}-\cdots$$
$$S = \sum\limits_{k=1}^{\infty}(-1)^{k+1}\frac{8k^2-1}{4k^2+2k} $$
$$S = \sum\limits_{k=1}^{\infty}(-1)^{k+1} \frac{(4k-1)(4k+1)}{2k(2k+1)}$$
