Science:Math Exam Resources/Courses/MATH101/April 2014/Question 09 (a)/Solution 1

We know that when a geometric series converges,

${\begin{aligned}{\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}.\end{aligned}}$

A geometric series converges when $-1<x<1$ which is what we have in the problem. If we differentiate both sides:

${\begin{aligned}{\frac {1}{(1-x)^{2}}}={\frac {d}{dx}}\left({\frac {1}{1-x}}\right)={\frac {d}{dx}}\sum _{n=0}^{\infty }x^{n}=\sum _{n=0}^{\infty }nx^{n-1}.\end{aligned}}$

To get the extra factor of $x$, multiply both sides by $x$:

${\begin{aligned}{\frac {x}{(1-x)^{2}}}=\sum _{n=0}^{\infty }nx^{n}.\end{aligned}}$

This multiplication is okay to do since the sum changes with $n$ and not $x$. This same logic is why we were not allowed to start by multiplying both sides by $n$.

We have shown the series relationship holds.