<s>[INST] Convergent value of $\sum\limits^\infty_{n=1}n\left(\frac{5}{6}\right)^{n-1}$ The sum in question is:
$$\sum^\infty_{n=1}n\left(\frac{5}{6}\right)^{n-1}$$
It passes the ratio test:
\begin{align}
&\lim_{n\rightarrow \infty}\frac{(n+1)\left(\frac{5}{6}\right)^{n}}{n\left(\frac{5}{6}\right)^{n-1}}\\
=\frac{5}{6}&\lim_{n\rightarrow \infty}\frac{(n+1)}{n}\frac{\left(\frac{5}{6}\right)^{n}}{\left(\frac{5}{6}\right)^{n}}\\
=\frac{5}{6}&\lim_{n\rightarrow \infty}(1+ \frac{1}{n})\\
=\frac{5}{6} &< 1\Rightarrow \text{convergent}
\end{align}
But now I do not know how to find the convergent value. 
 [/INST] $$f(x)= \frac{1}{1-x}=\sum_{n=0}^\infty x^n\\
f'(x)=\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}\\
f'(5/6)=\sum_{n=1}^\infty n(\frac{5}{6})^{n-1}= 36.$$
</s>