# 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.

• Do you have some specific reason for thinking it can be found in closed form rather than only numerically? (It's clear that it converges, so some numerical work should be able to give us a number as accurate as we need for any occasion.) ${}\qquad{}$ May 30 '15 at 17:32
• i don't think so if i can be found it in closed form .however it's clear that it is convergent serie May 30 '15 at 17:35
• Unhelpful comment: There is a closed form that I don't remember. May 30 '15 at 17:38
• $S=\log 27-\frac{\pi\sqrt3}{3}$
– Teoc
May 30 '15 at 17:40
• Googling "sum of reciprocals of pentagonal numbers" finds personal.psu.edu/jxs23/downey_ong_sellers_cmj_preprint.pdf and various other possibly useful references. May 30 '15 at 17:42

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.