How to sum $\frac{1}{9} + \frac{1}{18}+\frac{1}{30}+\frac{1}{45} + ......$ How to sum this series : 
$\frac{1}{9} + \frac{1}{18}+\frac{1}{30}+\frac{1}{45} + \frac{1}{65}......$
I am not getting any clue only a hint will be suffice please help. thanks.. 
 A: This series can be computed as $$\frac{1}{3}\sum_{k=0}^{\infty}\frac{1}{(\sum_{\ell=0}^k\ell)+2k+3}.$$
The partial sums are $$\frac{1}{3}\sum_{k=0}^{K}\frac{1}{(\sum_{\ell=0}^k\ell)+2k+3}=\frac{K+1}{3(K+3)},$$
and the limit is $1/3$. (Check for the first few values of $k$ and $K$ to convince yourself that the formulae are indeed correct and proceed by induction. Start with a value of $k=0$, not $1$. Hint: $\sum_{\ell=0}^k\ell=k(k+1)/2$.)
A: Observe
\begin{align*}
\frac{1}{9}+\frac{1}{18}+\frac{1}{30}+\frac{1}{45}+\ldots&=\frac{1}{3+3\cdot2}+\frac{1}{3+3\cdot2+3\cdot3}+\frac{1}{3+3\cdot2+3\cdot3+3\cdot4}+\ldots\\
&=\sum_{k=2}^{\infty}{\frac{1}{3\sum_{j=1}^{k}{j}}}\\
&=\sum_{k=2}^{\infty}{\frac{1}{3\left[\frac{k(k+1)}{2}\right]}}\\
&=\sum_{k=2}^{\infty}{\frac{2}{3\left[k(k+1)\right]}}\\
&=\frac{2}{3}\sum_{k=2}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}\right)\\
\end{align*}
Last one is a telescopic series, note we have
$$\frac{2}{3}\sum_{k=2}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}\right)=\frac{2}{3}\left(\frac{1}{2}\right)=\frac{1}{3}$$
