Infinite sum of reciprocals of pentagonal numbers How do I find this sum: $$\sum_{n=1}^\infty \frac{1}{p(n)}$$ where
$p(n)=\dfrac{n(3n-1)}{2}$ is the $n$th pentagonal number?
I know it is a convergent series, but I don't know if the sum can be found in closed form.
 A: Another way to do is just use basic calculus without using the digamma function: Let
$$ f(x)=\sum_{n=1}^\infty\frac{2}{n(3n-1)}x^{3n}. $$
Clearly $\sum_{n=1}^\infty\frac{2}{n(3n-1)}=f(1)$. Note
$$ f'(x)=6\sum_{n=1}^\infty\frac{1}{3n-1}x^{3n-1},f''(x)=6\sum_{n=1}^\infty x^{3n-2}=\frac{6x}{1-x^3}. $$
So
\begin{eqnarray}
f(1)&=&\int_0^1\int_0^x\frac{6t}{1-t^3}dtdx\\
&=&\int_0^1\int_t^1\frac{6t}{1-t^3}dxdt\\
&=&\int_0^1\frac{6t(1-t)}{1-t^3}dt\\
&=&\int_0^1\frac{6t}{1+t+t^2}dt\\
&=&\int_0^1\frac{6t}{(t+\frac{1}{2})^2+(\frac{\sqrt3}{2})^2}dt\\
&=&3\ln3-\frac{\pi}{\sqrt3}.
\end{eqnarray}
A: Using 
$$
   \frac{1}{p(n)} = 2 \left( \frac{1}{n-\tfrac{1}{3}} - \frac{1}{n} \right)
$$
and the definition of the digamma function:
$$
  \sum_{n=1}^\infty \frac{1}{p(n)} = -2 \left( \psi\left(\frac{2}{3}\right) + \gamma \right) = 3 \ln(3) - \frac{\pi}{\sqrt{3}} \approx 1.48204
$$
The value for the $\psi\left(\frac{2}{3}\right)$ can be derived from $\psi\left(\frac{1}{3}\right)$, states in the table of special values using the reflection identity.
