Computing $\sum_{n=1}^{\infty} \frac{\psi\left(\frac{n+1}{2}\right)}{ \binom{2n}{n}}$ Here is an interesting series I played with, namely 
$$\sum_{n=1}^{\infty} \frac{\displaystyle\psi\left(\frac{n+1}{2}\right)}{\displaystyle \binom{2n}{n}} \approx -0.245969181104090562617616399148$$
where $\psi(x)$ is digamma function
and the closed form I got you may see here, but that was possible because I had some luck.
However, the closed form doesn't look that friendly and maybe we can improve that.
It's also possible that a different approach leads to a shorter closed form, just the intuition.
I'd be interested in an approach that only uses series manipulations, I couldn't do that. 
 A: I took the expression from your paste-bin and simplified it quite a bit. I did it by first splitting the sum into its main parts and sorting them. A lot of the 201 terms looked similar enough so that I grouped them together. After a lot of simplifying and massaging one gets that 
$$sum = A + T + L + P$$
where
$$A = \frac{1}{135} \left(8 \sqrt{3} \pi -5 \gamma  \left(9+2 \sqrt{3} \pi \right)\right)$$
and
$$T = -\frac{2}{75} \left(50+13 \sqrt{5}\right) \pi  \tan ^{-1}\left(\sqrt{\frac{3}{5}}\right)+\frac{1}{75} \left(50+11 \sqrt{5}\right) \pi  \tan ^{-1}\left(\sqrt{15}\right)\\+\frac{1}{240} \left(125+29 \sqrt{5}\right)
   \sinh ^{-1}(2)-\frac{2 \text{csch}^{-1}(2)}{\sqrt{5}}+\frac{2 (5+\log (2)) \coth ^{-1}\left(\sqrt{5}\right)}{5 \sqrt{5}}$$
and
$$L = \frac{1}{40} \left(5-9 \sqrt{5}\right) \log (2)-\frac{\pi  \log (3)}{9 \sqrt{3}}+\log
   (4)+\frac{1}{80} \left(125+61 \sqrt{5}\right) \log \left(\sqrt{5}-1\right)\\+\frac{\log
   \left(\frac{1}{2} \left(3+\sqrt{5}\right)\right)}{\sqrt{5}}-\frac{\log (2) \left(800 \sqrt{3}
   \pi +9 \left(4585+645 \sqrt{5}-48 \sqrt{5} \log \left(\frac{1}{2}
   \left(3+\sqrt{5}\right)\right)\right)\right)}{10800}$$
and
$$P = \frac{4 \Re\left(\text{Li}_2\left(\frac{1}{8} \left(3+\sqrt{5}-i \sqrt{3} \left(-1+\sqrt{5}\right)\right)\right)-\text{Li}_2\left(\frac{1}{8} \left(3-\sqrt{5}+i \sqrt{6
   \left(3+\sqrt{5}\right)}\right)\right)\right)}{5 \sqrt{5}}\\-\frac{4 \Im\text{Li}_2\left(\frac{1}{6} \left(3-i \sqrt{3}\right)\right)}{3 \sqrt{3}}$$
Curiously
$$P - \gamma \approx 0.000220304\ldots$$
