<s>[INST] Prove the series $\sum_{n = 1}^{ \infty} \frac 1 6 n (\frac 5 6)^{n-1} = 6$ 
Prove the series $$\sum_{n = 1}^{ \infty} \frac 1 6 n (\frac 5 6)^{n-1} = 6.$$ 

I've tried various methods for proving the series:
The series is not geometric, but I see that $\frac 1 6 n (\frac 5 6)^{n-1} \rightarrow 0$. Also the series is not telescoping, as far I'm concerned.
What method could solve this ?
 [/INST] Here's a calculus approach:
We know that:
$$\frac{1}{1 - x} = 1 + x + x^2 + \dots$$
Differentiating both sides,
$$\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + \dots$$
Now, substitute $x = \frac{5}{6}$. Then,
$$\sum_{n = 1}^\infty n\left(\frac{5}{6}\right)^{n-1} = \frac{1}{\left(1 - \frac{5}{6}\right)^2}$$
So we have:
$$\frac{1}{6}\sum_{n = 1}^\infty n\left(\frac{5}{6}\right)^{n-1} = \frac{1}{6\left(1 - \frac{5}{6}\right)^2}\\
= 6$$
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