Summation of an infinite series The sum is as follows:
$$
\sum_{n=1}^{\infty} n \left ( \frac{1}{6}\right ) \left ( \frac{5}{6} \right )^{n-1}\\
$$
This is how I started:
$$
= \frac{1}{6}\sum_{n=1}^{\infty} n  \left ( \frac{5}{6} \right )^{n-1} \\
= \frac{1}{5}\sum_{n=1}^{\infty} n  \left ( \frac{5}{6} \right )^{n}\\\\
 = \frac{1}{5}S\\
S = \frac{5}{6} + 2\left (\frac{5}{6}\right)^2 + 3\left (\frac{5}{6}\right)^3 + ...
$$
I don't know how to group these in to partial sums and get the result. I also tried considering it as a finite sum (sum from 1 to n) and applying the limit, but that it didn't get me anywhere!
PS: I am not looking for the calculus method.

I tried to do it directly in the form of the accepted answer,
$$
\textrm{if} \ x= \frac{5}{6},\\
S = x + 2x^2 + 3x^3 + ...\\
Sx = x^2 + 2x^3 + 3x^4 + ...\\
S(1-x) = x + x^2 + x^3 + ...\\
\textrm{for x < 1},\ \  \sum_{n=1}^{\infty}x^n = -\frac{x}{x-1}\  (\textrm{I looked up this eqn})\\
S = \frac{x}{(1-x)^2}\\
\therefore S = 30\\
\textrm{Hence the sum} \sum_{n=1}^{\infty} n \left ( \frac{1}{6}\right ) \left ( \frac{5}{6} \right )^{n-1} = \frac{30}{5} = 6 
$$

hint: differentiate  the identity
$$\sum_{k=0}^{\infty} x^k = \frac{1}{1-x} $$
