<s>[INST] Find sum: $\sum_{n=1}^\infty \frac{3^{n+1}}{6^n + 1} $ I'm a little confused about how to find sums. I know that to find the sum, we need to take $\lim_{n\ \to \infty}  s_n$, where $s_n$ is the partial sum. This is confusing for me b/c I have trouble telling the difference between $s_n$ and $a_n$, where $a_n$ is the series formula in $\sum a_n$. 
So anyway, this is my approach: 
$$s_n = \sum_{i=1}^n \frac{3^{i+1}}{6^i+1} = \frac{9}{7} + \frac{27}{37} + \frac{81}{217} + ...+ \frac{3^{n+1}}{6^n+1}$$
See, I know ultimately that I have to take the limit of $s_n$, however any convergent series (I know this converges) will have a limit = 0. So i'm assuming that in order to find the sum I need to manipulate the sum formula. Is this the right approach?
Thanks!
 [/INST] $$
\frac{3^n}{6^n+1}=\frac{3^n}{6^n(1+6^{-n})}=\frac{1}{2^n(1+6^{-n})}=\frac{A}{2^n}-\frac{B}{1+6^{-n}}=\frac{1}{2^n}-\frac{12^{-n}}{1+6^{-n}}= \\[6mm]
\frac{1}{2^n}-\frac{1}{6^n(6^n+1)}=\frac{1}{2^n}-\frac{1}{6^n}+\frac{1}{6^n+1}\quad\Rightarrow\quad\sum_{n=1}^{\infty}\frac{3^{n+1}}{6^n+1}=\\[6mm]
3\sum_{n=1}^{\infty}\left(\frac{1}{2^n}-\frac{1}{6^n}+\frac{1}{6^n+1}\right)=\color{red}{3\left(1-\frac{1}{5}+\frac{1}{\log6}\left[\varPsi_{1/6}\left(1-\frac{i\,\pi}{\log6}\right)-\log\frac65\right]\right)} \\[8mm]
\small\qquad\qquad\qquad\qquad\qquad\qquad\qquad\varPsi_{q}(z)\quad\text{q-Polygamma Function}
$$
</s>