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.

We have:
$$ \sum_{m\geq 0}\frac{1}{(m+a)(m+b)}=\frac{\psi(a)-\psi(b)}{a-b} \tag{1}$$
hence it is enough to compute:
$$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\sum_{m\geq 0}\frac{1}{\left(m+\frac{n}{2}\right)\left(m+\frac{n+1}{2}\right)}\tag{2} $$
or:
$$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\sum_{m\geq 0}\int_{0}^{+\infty}e^{-\left(m+\frac{n}{2}\right)x}-e^{-\left(m+\frac{n+1}{2}\right)x}\,dx\tag{3} $$
or:
$$ \sum_{m\geq 0}\int_{0}^{+\infty}\log(1+e^{-x/2})e^{-(m+1/2)x}(e^{x/2}-1)\,dx \tag{4}$$
that is equivalent to:
$$ \int_{0}^{+\infty}\frac{e^{x/2}\log(1+e^{-x/2})}{1+e^{x/2}}\,dx = \zeta(2)-\log^2(2) \tag{5}$$
through dilogarithms, to have (for the original question):
$$ \sum_{n\geq 1}\frac{(-1)^n}{n}\left(\psi\left(\frac{n}{2}\right)-\psi\left(\frac{n+1}{2}\right)\right)= \color{red}{\zeta(2)-\log^2(2)}.\tag{6}$$
In the same spirit,

$$\begin{eqnarray*}\sum_{n\geq 1}(-1)^{n+1}\left(\psi\left(\frac{n}{2}\right)-\psi\left(\frac{n+1}{2}\right)\right)&=&\frac{1}{2}\sum_{n\geq 1}\sum_{m\geq 0}\frac{(-1)^n}{\left(m+\frac{n}{2}\right)\left(m+\frac{n+1}{2}\right)}\\[0.2cm]&=&2\sum_{n\geq 1}\sum_{m\geq 0}\left(\frac{(-1)^n}{2m+n}-\frac{(-1)^n}{2m+n+1}\right)\\[0.2cm]&=&-2\sum_{m\geq 0}\int_{0}^{+\infty}e^{-(2m+1)x}\tanh(x/2)\,dx\\[0.2cm]&=&-2\int_{0}^{+\infty}\frac{e^x}{(1+e^x)^2}\,dx\\[0.2cm]&=&\color{red}{-1}.\tag{7}\end{eqnarray*}$$

