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

Start with the formula we derived in (a). If we differentiate this formula with respect to $x$,

${\begin{aligned}\sum _{n=0}^{\infty }n^{2}x^{n-1}&={\frac {d}{dx}}\sum _{n=0}^{\infty }nx^{n}\\&={\frac {d}{dx}}\left({\frac {x}{(1-x)^{2}}}\right)\\&={\frac {1}{(1-x)^{2}}}+{\frac {2x}{(1-x)^{3}}}\\&={\frac {1+x}{(1-x)^{3}}}.\end{aligned}}$

If we multiply both sides by $x$:

${\begin{aligned}\sum _{n=0}^{\infty }n^{2}x^{n}={\frac {x(1+x)}{(1-x)^{3}}}\end{aligned}}$

then we have the series we are interested in. We note that the two expression are only equivalent if the series converges. We will check the radius of convergence by using the ratio test.

${\begin{aligned}\lim _{n\to \infty }\left|{\frac {(n+1)^{2}x^{n}}{n^{2}x^{n-1}}}\right|=\left(\lim _{n\to \infty }\left|{\frac {(n+1)^{2}}{n^{2}}}\right|\right)|x|=|x|.\end{aligned}}$

By the ratio test, if $|x|<1$, the series converges and if $|x|>1$, the series diverges. This is unsurprising since we already know this is required for the geometric series in (a). We have to check the boundaries independently. If $x=1$, the expression

${\frac {x(1+x)}{(1-x)^{3}}}$

fails to exist and so the series must diverge. If $x=-1$, this expression is finite but that does not mean the series converges. We have to look at the series itself which becomes

${\begin{aligned}\sum _{n=0}^{\infty }(-1)^{n}n^{2}&=-1^{2}+2^{2}-3^{2}+4^{2}-5^{2}+6^{2}+\cdots =(-1^{2}+2^{2})+(-3^{2}+4^{2})+(-5^{2}+6^{2})+\cdots \\&=(2-1)(1+2)+(3-2)(3+4)+(6-5)(5+6)+\cdots =\sum _{n=0}^{\infty }n,\end{aligned}}$

and hence diverges. Thus, the series only converges for $x\in (-1,1)$.