# Sum involving the Digamma function.

Consider the double sum $$f(x):=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{1}{m+\frac{1}{n+x}}-\frac{1}{m+\frac{1}{n}},$$ which converges for $x>0$. One interpretation of this sum is the measure of the double preimage $T^{-2}([0,x])$, in which $T:[0,1]\to [0,1]$ denotes the Gauss map $T(x)=\{1/x\}$ (the fractional part of $1/x$). Let $\psi$ denote the Digamma function $\psi(z):=\Gamma'(z)/\Gamma(z)$. The identity $$\psi(z)=-\gamma+\sum_{k=0}^\infty\left(\frac{1}{n+1}-\frac{1}{n+z}\right)$$ can then be used to rewrite $f$ as $$f(x)=\sum_{n=1}^\infty \psi\left(1+\frac{1}{n}\right)-\psi\left(1+\frac{1}{n+x}\right).$$ Is there any closed-form simplification of the sum above? It seems plausible that there may be such a solution involving the higher Polygamma function $\psi^{(2)}$.