Summation of a multiplication How can I prove $$\sum_{k=1}^\infty kq^{k-1}=\frac{1}{(1-q)^2}$$?
I know that the formula to do the summation of a number $a$ is $a(a+1)/2$. And I also know that the summation of a geometric series $ar^n$ is $a/(1-r)$. 
But I don't know how to solve my problem.
 A: We have
$$\frac{1}{1-q}=1+q+q^2+\dots$$
The sequence you have is then
$1+q+q^2+q^3+\dots$
$\;\;\;\;+q+q^2+q^3+\dots$
$\;\;\;\;\;\;\;\;\;\;\;+q^2+q^3+\dots$ 
which equals
$$(1+q+q^2+\dots)+q(1+q+q^2+\dots)+q^2(1+q+q^2+\dots)+\dots$$
$$=(1+q+q^2+\dots)(1+q+q^2+\dots)$$
$$=\dfrac{1}{(1-q)^2}$$
A: You'll realise on expansion of a few terms that it forms an AGP. (Read more about AGP elsewhere; it is when an AP and a GP reside in the same term of a series.) The formula for sum of the following AGP is
$a+(a+d)r+(a+2d)r^2+\cdots+(a+(n-1)d)r^{n-1}$
($a, d, r$ have their usual meanings in context of sequences and series)
$S_n=\frac{a}{1-r}+\frac{dr}{(1-r)^2}-\frac{dr^{n}}{(1-r)^2}-\frac{[a+(n-1)d]r^n}{1-r}$
For what you want to prove, $n=∞$, hence, any term with $r^n$ disappears (note how this implies $q<1$ because $r=q$) The following are the values for what you are trying to prove. $a=1, r=q, d=1$ 
Plug 'em in and off you pop.
$S_{∞} = \frac{1}{1-q}+\frac{q}{(1-q)^2} = \frac{1}{(1-q)^2}$
P.S, if you want to derive the formula, and you should, as a little exercise, flip the series, move it forward by one, multiply by $r$ and subtract it from the original series in order. All the terms should cancel out except the ones that make up the formula.
