Infinite sum of reciprocals of pentagonal numbers

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}

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.