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,

$$
\textrm{if} \ x= \frac{5}{6},\\
S = x + 2x^2 + 3x^3 + ...\\
Sx = x^2 + 2x^3 + 3x^4 + ...\\
S(1-x) = x + x^2 + x^3 + ...\\
\textrm{for x &lt; 1},\ \  \sum_{n=1}^{\infty}x^n = -\frac{x}{x-1}\  (\textrm{I looked up this eqn})\\
S = \frac{x}{(1-x)^2}\\
\therefore S = 30\\
\textrm{Hence the sum} \sum_{n=1}^{\infty} n \left ( \frac{1}{6}\right ) \left ( \frac{5}{6} \right )^{n-1} = \frac{30}{5} = 6 
$$
Letting $a = d = 1/6$ and $r = 5/6$, our sum is:
$$
S = a + (a + d)r + (a + 2d)r^2 + (a + 3d)r^3 + \cdots
$$
Scaling by $r$, we find that:
$$
rS = ar + (a + d)r^2 + (a + 2d)r^3 + \cdots
$$
Subtracting the two equations (by collecting like powers of $r$), we obtain:
$$
(1 - r)S = a + dr + dr^2 + dr^3 + \cdots = a + dr(1 + r + r^2 + \cdots) = a + \frac{dr}{1 - r}
$$
Hence, we conclude that:
$$
S = \frac{a}{1 - r} + \frac{dr}{(1 - r)^2} = \frac{1/6}{1 - 5/6} + \frac{(1/6)(5/6)}{(1 - 5/6)^2} = \frac{1}{6 - 5} + \frac{(1)(5)}{(6 - 5)^2} = 6
$$
hint: differentiate  the identity

$$\sum_{k=0}^{\infty} x^k = \frac{1}{1-x} $$