Series with digammas (Inspired by a comment in answer https://math.stackexchange.com/a/699264/442.)  
corrected 
Let $\Psi(x) = \Gamma'(x)/\Gamma(x)$ be the digamma function.  Show
$$
\sum_{n=1}^\infty (-1)^n\left(\Psi\left(\frac{n+1}{2}\right)
-\Psi\left(\frac{n}{2}\right)\right) = -1
$$
As noted, it agrees to many decimals.  But care may be required since the convergence is only conditional.
added
Both solutions are good.  But could use explanation for exchange of summation.
Note that
$$
2\sum_{n=1}^\infty\sum_{m=0}^\infty \left(\frac{(-1)^n}{2m+n}
-\frac{(-1)^n}{2m+n+1}\right)
=
2\sum_{m=0}^\infty\sum_{n=1}^\infty \left(\frac{(-1)^n}{2m+n}
-\frac{(-1)^n}{2m+n+1}\right)
= -1
$$
is correct.  But "Fibini" justification fails, since
$$
\sum_{m=0}^\infty\sum_{n=1}^\infty \left|\frac{(-1)^n}{2m+n}
-\frac{(-1)^n}{2m+n+1}\right| = \infty
$$
Similarly in Random Variable's solution, the exchange $\sum_{n=1}^\infty \int_0^1 = \int_0^1 \sum_{n=1}^\infty$, although correct, cannot be justified by Fubini.

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The Digamma Function has an interesting identity:
  $\ds{\Psi\pars{z + 1 \over 2} - \Psi\pars{z \over 2} = 2\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over k + z}}$ such that

\begin{align}
&\color{#f00}{\sum_{n = 1}^{\infty}\pars{-1}^{n}
\bracks{\Psi\pars{n + 1 \over 2} -\Psi\pars{n \over 2}}} =
\sum_{n = 1}^{\infty}\pars{-1}^{n}\bracks{%
2\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over k + n}}
\\[3mm] = &\
2\sum_{n = 1}^{\infty}\pars{-1}^{n}
\sum_{k = 0}^{\infty}\pars{-1}^{k}\int_{0}^{1}x^{k + n - 1}\,\dd x =
-2\int_{0}^{1}\sum_{n = 1}^{\infty}\pars{-x}^{n - 1}
\sum_{k = 0}^{\infty}\pars{-x}^{k}\,\dd x
\\[3mm] = &\
-2\int_{0}^{1}{1 \over 1 + x}\,{1 \over 1 + x}\,\dd x = \color{#f00}{-1}
\end{align}
~     
