So, simply, where do I go wrong?

The question:

The interval of the power series$\displaystyle \sum ^\infty _{n=1}\frac{(ax)^n}{n^2}$ is $\displaystyle \left[ -\frac{1}{3},\frac{1}{3} \right] $. Find $\displaystyle a$.

My solution:

For the series to be convergent, d'Alemberts ratio test must be satisfied, i.e. $\displaystyle \lim_{n\rightarrow \infty} |\frac{u_{n+1}}{u_n}|<1$ must hold true.

For the series given we have:

$\displaystyle \lim_{n\rightarrow \infty} |\frac{u_{n+1}}{u_n}|=\lim_{n\rightarrow \infty} |\frac{(ax)^{n+1}n^2}{(n+1)^2(ax)^n}| $

$\displaystyle = \lim_{n\rightarrow\infty}\frac{n^2}{(n+1)^2}|ax|=| ax|=|a||x|<1\Leftrightarrow |x|<\frac{1}{|a|} $.

This gives $\displaystyle -\frac{1}{|a|}<x<\frac{1}{|a|}$

It is easy to show that the series is convergent also for $\displaystyle x=\frac{1}{|a|}$ and $\displaystyle x=-\frac{1}{|a|}$ (by making these substitutions into the series given).

So now we have that $\displaystyle -\frac{1}{|a|}\leq x \leq \frac{1}{|a|}$ for the series to be convergent.

But the interval of convergence is $\displaystyle \left[ -\frac{1}{3},\frac{1}{3} \right] $, meaning that $\displaystyle a= \pm 3$.

.

However, the key in my book claims the answer to be $\displaystyle a=\frac{1}{3}$

Thank you in advance

Aliquantus