<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] Recall the geometric sum:
$$\sum_{k=0}^\infty x^n = \frac{1}{1-x}$$
Take the derivative of both sides:
$$\sum_{k=1}^\infty nx^{n-1} = \frac{1}{(1-x)^2}$$
Your sum appears when $x=5/6$. The above sums only hold if $|x|<1$ which does hold here.
</s>