Here is an interesting formula for the reciprocal of the heptagonal numbers. Are there any other analogous formulas for the polygonal or figurate numbers?
$$ \sum_{n=1}^\infty \frac{2}{n(5n-3)} =\\ \frac{1}{15}{\pi}{\sqrt{25-10\sqrt{5}}}+\frac{2}{3}\ln(5)+\frac{{1}+\sqrt{5}}{3}\ln\left(\frac{1}{2}\sqrt{10-2\sqrt{5}}\right)+\frac{{1}-\sqrt{5}}{3}\ln\left(\frac{1}{2}\sqrt{10+2\sqrt{5}}\right) $$
Reference: "Beyond The Basel Problem" L. Downey ,B. Ong ,J. Sellers
check this out :
http://en.wikipedia.org/wiki/Polygonal_number
it's not hard to prove these formulas
figurate numbers
$$ \sum_{n=1}^{\infty} \frac{1}{\dbinom{n+r-1}{r}} = \sum_{n=1}^{\infty} \frac{r!(n-1)!}{(n+r-1)!} $$
$$ = r\sum_{n=1}^{\infty} \frac{\Gamma(r) \Gamma(n)}{\Gamma(n +r )} = r\sum_{n=1}^{\infty} \beta(r,n) = r\sum_{n=1}^{\infty} \int_0^1 x^{n-1}(1 - x)^{r-1} \ dx = r\int_0^1 (1-x)^{r-1} \sum_{n=1}^{\infty} x^{n-1} \ dx $$
$$ = r\int_0^1 (1-x)^{r-2} \ dx = -r\left[ \frac{(1-x)^{r-1}}{r-1} \right]_0^1 = \frac{r}{r-1} $$