Horrible infinite sum of reciprocals

Note that $\sum_{n=0}^N \frac{1}{n+z} \; = \; \sum_{n=1}^{N+1}\frac{1}{n} - \gamma - \psi_N(z)$ where $\psi_N(z) \to \psi(z)$ as $N \to \infty$ . Here $\psi$ is the digamma function. This result enables us to evaluate the two infinite sums.

$\begin{aligned} X & = \; \sum_{n=0}^\infty\left(\frac{1}{3n+1} - \frac{1}{3n+2}\right) \; = \; - \tfrac13\big(\psi(\tfrac13) - \psi(\tfrac23)\big) \; = \; \tfrac13\pi\cot\tfrac13\pi \\ & = \; \frac{\pi}{3\sqrt{3}} \end{aligned}$ Similarly $\begin{aligned} Y & = \; \sum_{n=0}^\infty\left(\frac{1}{6n+1} - \frac{1}{6n+3} - \frac{1}{6n+4} + \frac{1}{6n+6}\right) \\ & = \; -\tfrac16\left(\psi(\tfrac16) - \psi(\tfrac12) - \psi(\tfrac23) + \psi(1)\right) \\ & = \; \frac{\pi}{3\sqrt{3}} \end{aligned}$ after substituting in standard values for the digamma function. Thus the value of the stated expression is $X - Y = 0 = \sqrt{0} + \ln1$ , making the answer $0 + 1 = \boxed{1}$ .