<s>[INST] Finding out $S:=1+\frac12-\frac13-\frac14+\frac15+\frac16-\frac17-\frac18+\cdots$ I was willing to determine the sum of following 
$$S:=1+\frac12-\frac13-\frac14+\frac15+\frac16-\frac17-\frac18+\cdots$$
I tried the following 
\begin{align*}
S=&\sum\limits_{n=0}^\infty (-1)^n\left(\frac{1}{2n-1}+\frac{1}{2n}\right)\\
 =&\sum\limits_{n=0}^\infty (-1)^n\left(\int_0^1 x^{2n-2}dx+\int_0^1 x^{2n-1} dx\right)\\
=&\int_0^1 \sum\limits_{n=0}^\infty (-1)^n\left(x^{2n-2}+x^{2n-1}\right)\\
=&\int_0^1 [x^{-2} \sum\limits_{n=0}^\infty (-1)^nx^{2n}+x^{-1} \sum\limits_{n=0}^\infty (-1)^nx^{2n}]
\end{align*}
and don't know after his what to do . Can you please help me on this regard?
Thanking you in advance
 [/INST] Let $$S = \sum_{n = 1}^{\infty} (-1)^n\left(\frac{1}{2n-1}+\frac{1}{2n}\right).$$ Then
$\begin {eqnarray}
S & = & \sum_{k = 1}^{\infty} (-1)^{2k}\left(\frac{1}{4k-1}+\frac{1}{4k}\right) + \sum_{k = 1}^{\infty} (-1)^{2k+1}\left(\frac{1}{4k+1}+\frac{1}{4k+2}\right) \nonumber \\ & = & \sum_{k = 1}^{\infty} \left(\frac {1} {4k - 1} - \frac {1} {4k + 1}\right) + \sum_{k =1}^{\infty} \left(\frac {1} {4k} - \frac {1} {4k + 2}\right),
\end {eqnarray}$
which is easier to proceed now.
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