How to solve the equation $\sum_{n=1}^\infty n^{-2}/\binom{n+x}{n} =\frac{3}{2}$ for $x$? I find this problem on this page.
Find $x\in\mathbb{R}$ such that 
$$\sum_{n=1}^\infty \frac{1}{n^2\displaystyle\binom{n+x}{n}}=\frac{3}{2}$$
Thank you very much.
 A: Writing the binomial coefficient as a beta function and then expressing this beta function in terms of its integral representation (Euler), we can find out what function the series evaluates to for general $x$, with $x>-1$:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{1}{n^2\binom{n+x}{n}}
&=\sum_{n=1}^{\infty}\frac{\operatorname{B}{\left(n,x+1\right)}}{n}\\
&=\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{1}t^{n-1}\left(1-t\right)^{x}\,\mathrm{d}t\\
&=\int_{0}^{1}\left[\sum_{n=1}^{\infty}\frac{t^{n-1}}{n}\right]\left(1-t\right)^{x}\,\mathrm{d}t\\
&=\int_{0}^{1}\left[-\frac{\ln{\left(1-t\right)}}{t}\right]\left(1-t\right)^{x}\,\mathrm{d}t\\
&=-\int_{0}^{1}\frac{\left(1-t\right)^{x}\ln{\left(1-t\right)}}{t}\,\mathrm{d}t\\
&=-\int_{0}^{1}\frac{z^{x}\ln{\left(z\right)}}{1-z}\,\mathrm{d}z\\
&=\frac{\partial}{\partial x}\int_{0}^{1}\frac{1-z^{x}}{1-z}\,\mathrm{d}z\\
&=\frac{\partial}{\partial x}\left[\gamma+\psi{\left(x+1\right)}\right]\\
&=\psi^{(1)}{\left(x+1\right)},\\
\end{align}$$
where $\psi{\left(x\right)}$ and $\psi^{(1)}{\left(x\right)}$ are, respectively, the digamma function and the trigamma function. In other words, we must solve the equation $\psi^{(1)}{\left(x+1\right)}=\frac32$. Numerically, the solution $x$ has an approximate value,
$$x\approx 0.0656560393745264854445863686\dots,$$
but I'm very pessimistic about our chances of finding a closed form for this value. In particular $x$ is not a rational number.

Edit:
After encountering the closed form of a possibly related series proposed by Chris'sis in this question, I'm about ready to admit that anything is possible. At the very least, I'm much less confident that the possible existence of a closed form can be altogether ruled out.
