Radius of convergence of $\sum_{n\ge 1}{\frac{x^{n^2}}{n^2}}$

Asked Nov 17 '13 at 06:03

Active Nov 17 '13 at 06:10

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I want to determine the radius of convergence of the power series $$\sum_{n\ge 1}{\frac{x^{n^2}}{n^2}}$$

Is my following try correct, and is there any simpler way to do this:

Put $a_n(x)=\frac{x^{n^2}}{n^2}$. If $x\not = 0$, $$\left|\frac{a_{n+1}(x)}{a_n(x)}\right|=\frac{n^2x^{2n+1}}{(n+1)^2}\sim_\infty x^{2n+1}$$

Now $ x^{2n+1}$ goes to $0<1$ when $|x|<1$ and goes to $\infty$ otherwise. Hence the radius of convergence is $R=1$.

edited Nov 17 '13 at 06:10

2'5 9'2

asked Nov 17 '13 at 06:03

palio

5

$$\text{Yes!}$$ – meta_warrior Nov 17 '13 at 06:05
1

Additionally, this series in fact converges for $x=\pm 1$. Its value at $x=1$ is $\frac{\pi^2}{6}$ and its value at $x=-1$ is $-\frac{\pi^2}{12}$. – Cameron L. Williams Nov 17 '13 at 06:06
If you are interested in seeing this series for complex $x$, see http://math.stackexchange.com/questions/84779/fractal-behavior-along-the-boundary-of-convergence – 2'5 9'2 Nov 17 '13 at 06:09
@alex.jordan Ugh. I had flashbacks from when I did this problem in my complex analysis course. I misinterpreted the exponent entirely (I thought it said $z^{2n}$...) and got the problem completely wrong. – Cameron L. Williams Nov 17 '13 at 06:10

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