Can anyone help me understand how to solve these two series? More than the solution I'm interested in understanding which process I should follow.
$$ \sum_{i = 3}^{\infty} i * a^{i-1}, 0 < a < 1. $$
$$ \sum_{i = 3}^{\infty} i\sum_{k = 2}^{i-1} a^{i-k} * b^{k-2} , 0 < a < 1, 0 < b < 1. $$
These two series come as part of a long mathematical proof which I omitted for brevity, if you think it is relevant I will post it.
On
For the first one, recall that $$\frac{1}{1 - x} = \sum\limits_{n = 0}^{\infty} x^n = 1 + x + x^2 + ...$$
Differentiating both sides leads to
$$\frac{1}{(1 - x)^2} = \sum\limits_{n = 1}^{\infty} = 1 + 2x + 3x^2 + ... = \sum\limits_{n = 1}^{\infty} nx^{n - 1}$$
This is valid within the disk of convergence, which has radius $1$ around $x = 0$.
Hint: For the first one if $\sum_{i = 3}^{\infty} x^{i}=f(x)$ then $\sum_{i = 3}^{\infty} i \times x^{i-1}=f'(x)$.
For the second one consider that: $$ \sum_{k = 2}^{i-1} a^{i-k} b^{k-2}=a^{i-2}\frac{1-\frac{b^{i-2}}{a^{i-2}}}{1-\frac{b}{a}}= \frac{a^{i-2}-{b^{i-2}}}{1-\frac{b}{a}} $$ Then you can decompose the series into two one and use the previous step.