# Radius and interval of convergence for $\sum \limits_{n=1}^\infty \frac{x^{n^2}}{2^n}$ [closed]

$\sum \limits_{n=1}^\infty \frac{x^{n^2}}{2^n}$

I shall find the series radius and interval of convergence. I tried to use the ratio test and the nth root test to check the absolute convergence but failed. Please give me a hint how to approach this series.

## closed as off-topic by Namaste, Saad, Xander Henderson, user284331, Chris CusterApr 10 '18 at 0:40

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• Some hints: We have $|x^{n^2}| \leq |x|^n$ for $|x|\leq 1$. If $x>1$ then $x^{n^2} = (x^n)^n$ and for large $n$ we will have $|x|^n > 2$. – Winther Mar 4 '18 at 20:54
• Try the ratio test again, or explain where you got stuck. – Matthew Leingang Mar 4 '18 at 20:54

For the ratio test, we have \begin{align*} \lim_{n\to\infty}\left\lvert\frac{x^{(n+1)^2}}{2^{n+1}}\cdot\frac{2^n}{x^{n^2}}\right\rvert&=\lim_{n\to\infty}\frac{\lvert x\rvert^{2n+1}}{2}. \end{align*} Notably, this limit is $0$ if $\lvert x\rvert<1$, $\frac{1}{2}$ if $\lvert x\rvert=1$, and positive infinity if $\lvert x\rvert>1$. What does that tell you?
The coefficients of this power series are $$a_k=\begin{cases}\frac{1}{2^n}, &\text{if }k=n^2\text{ for some }n\\ 0, &\text{else}\end{cases}.$$ So $$\limsup_{k\to\infty}|a_k|^{\frac{1}{k}}=\limsup_{n\to\infty}\left|\frac{1}{2^n}\right|^{\frac{1}{n^2}}=\limsup_{n\to\infty}\left(\frac{1}{2}\right)^{\frac{1}{n}}=1.$$ Thus the radius of convergence is $1$. Also it's trivial that the series converges if $|x|=1$. Hence the interval of convergence is $[-1,1]$.