<s>[INST] Calculus - Help with this series $S_n=\sum_{k=1}^\infty \frac{1}{9k^2+3k-2}$ $$S_n=\sum_{k=1}^\infty \frac{1}{9k^2+3k-2}$$
My solution
$$S_n=\sum_{k=1}^\infty \frac{1}{9k^2+3k-2} = \sum_{k=1}^\infty \frac{1}{3k(3k+1)-2}=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{3k(3k+1)-2}$$
I understand that it converge (since r<1), but I need to find the sum.
Any ideas?
Thanks!
 [/INST] As has been pointed out in hints the following holds:
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{9k^2 + 3k-2} &= \frac{1}{3} \sum_{n=1}^{\infty} \left(\frac{1}{3k-1}-\frac{1}{3k+2}\right) \\
&= \frac{1}{3} \left[ \frac{1}{2} + \sum_{n=1}^{\infty} \frac{1}{3n+2} - \sum_{k=1}^{\infty} \frac{1}{3k+2} \right] \\
&= \frac{1}{6}.
\end{align}
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