<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] Hmm so write $f(x) = \sum_{n=1}^{\infty}x^n$. We know how to write this as a geometric series sum, and know the answer is $\frac{x}{1-x}$ (whenever $|x|<1$, which is the case with $x=5/6$ fortunately.
Now we will differentiate $f(x)$ and set $x=5/6$. We are allowed to do this by Taylor's theorem and term by term differentiation, because our series is just a Taylor series for $f(x)$.
If we differentiate the expression $\frac{x}{1-x}$ in $x$ and set $x=5/6$ we will get the answer.
</s>