What does the geometric series $ \sum\limits_{n=0}^{\infty} n \cdot \left( \frac{1}{4} \right) ^{n}$ converge to? I am trying to find the value to which the geometric series
$$
\sum\limits_{n=0}^{\infty} n \cdot \left( \frac{1}{4} \right) ^{n},
$$
converges. Now I know how to find a sum when we have a common ratio, but since $n$ is increased with each term, I am stuck at how to tackle this.
Any suggestions would be appreciated.

Hint
$$S_n=1\cdot\left( \frac{1}{4} \right) ^{1}+2\cdot\left( \frac{1}{4} \right) ^{2}+...+n\cdot\left( \frac{1}{4} \right) ^{n}$$
Now, multiply by $1/4$
$$\frac{1}{4}S_n=1\cdot\left( \frac{1}{4} \right) ^{2}+2\cdot\left( \frac{1}{4} \right) ^{3}+...+n\cdot\left( \frac{1}{4} \right) ^{n+1}$$
Now subtract,
$$S_n-\frac{1}{4}S_n=1\cdot\left( \frac{1}{4} \right) ^{2}+(2-1)\cdot\left( \frac{1}{4} \right) ^{2}+...+(n-(n-1))\left( \frac{1}{4} \right) ^{n}+n\cdot\left( \frac{1}{4} \right) ^{n+1}$$
Did you understand the idea? Can you finish?
