Radius of convergence of $\sum_{n=0}^{\infty } \frac{x^{n^{2}}}{2^{n}}$ $$\sum_{n=0}^{\infty } \frac{x^{n^{2}}}{2^{n}}$$
How to find the' radius and interval of convergence and identify the values of $x$ for which the series converges?
 A: For $x\leq 1$ you can take by comparism 
$$ \sum_{n=0}^\infty \left| \frac{x^{n^2}}{2^n } \right| \leq \sum_{n=0}^\infty \frac{1}{2^n}$$
As $2^n \leq n!$ the series of one of you last threads is a smaller divergent sum. 
So you see the radius of convergence is 1.
A: Of course, this converges for $x=0$.
Now for $x\neq 0$, use the ratio test, and consider
$$
\frac{x^{(n+1)^2}2^n}{2^{n+1}x^{n^2}}=\frac{x^{2n+1}}{2}.
$$
This proves convergence when $0<|x|\leq 1$ (limit is $0$ or $1/2$, hence $<1$)and divergence when $|x|>1$ (limit is $+\infty$).
So the radius is $1$ and the interval is $[-1,1]$.
A: Clearly ,if $|x|<1$ The series converges(Comparison with geometric series).
If $|x|>1 $ then $|\frac{x^{n^2}}{2^n}|= 2^{n^2log_2x-n}$ which grows arbitrarily large.
A: Hint only: Use the Ratio test to get started:
$$
\lim_{n\to \infty} \frac{a_{n+1}}{a_n} = \lim_{n\to \infty}\frac{x^{(n+1)^2}2^n}{x^{n^2}2^{n+1}} = \lim_{n\to \infty} x^{n^2 + 2n + 1 - n^2}\frac{1}{2}.
$$
A: Go back to the definition: it's the supremum of the number $r>0$ such that $\{|a_n|r^n\}$ is bounded. 
If $\{x^{n^2}2^{-n}\}$ is bounded, so is $\{x^n\}$ which implies that the radius of convergence is smallet than $1$. As the series is convergent for $|x|=1$, actually there is normal convergence of the power series on the whole closed disk. 
A: Proceeding according to the rules of the book we have to compute the $\limsup_{m\to\infty}\root m\of{|a_m|}$. Since
$$|a_m|=\cases{2^{-n}\quad&$(m=n^2,\ n\geq0)$ \cr 0&(otherwise)\cr}$$
it follows that
$$\limsup_{m\to\infty}\root m\of{|a_m|}=\limsup_{n\to\infty}\bigl(2^{-n}\bigr)^{1/n^2}=\lim_{n\to\infty}\root n\of{1/2}=1\ .$$
The radius of convergence $\rho$ is the reciprocal of this limit, so  $\rho=1$. In the case at hand the series obviously converges for all complex $x$ with $|x|\leq1$, and it will diverge when $|x|>1$.
