To find radius of convergence of $\sum_{n=0}^{\infty}2^{2n}x^{n^{2}}$? How to find radius of convergence of the  power series $\displaystyle\sum_{n=0}^{\infty}2^{2n}x^{n^{2}}$? That raised to $n^2$ part confusing me.Can I do this? Instead of taking $\frac{1}{n}$th root can I take $\frac{1}{n^{2}}$ root for applying root test? 
 A: Your series is 
$$\sum_{n=1}^\infty a_n x^n\;\;,\;\;a_n=\begin{cases}0,&n\text{ is not a square}\\{}\\2^{2k},&n=k^2\end{cases}$$
Thus, by the Cauchy-Hadamard Formula
$$\frac1R=\limsup_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\left(2^{2\sqrt n}\right)^{1/n}=\lim_{n\to\infty}4^{1/\sqrt n}=1$$
and thus the radius of convergence is $\;R=1\;$ .
The answer's already been given, yet your series, as it is treated and as you wrote, is not a power series as formally defined. Thus, it is probably that in this exercise you should first try to present the series as a powers one and then find its convergence radius.
A: Note that 
$$\sqrt[n]{|2^{2n}x^{n^2}|}=4|x|^n$$
For $|x|<1$, $\limsup_{n\to\infty}4|x|^n=0$ and for $|x|>1$, $\limsup_{n\to\infty}4|x|^n=\infty$.
Inasmuch as the series of interest diverges if $|x|=1$, we find that the series of interest converges if and only if $|x|<1$.  
A: Use the root test.  You have
$$\left|2^{2n} x^{n^2}\right|^{1/n} = 4|x|^n.$$
This quantity goes to zero if $|x| < 1$ and fails the root test otherwise.
Therefore the series converges absolutely for $|x| < 1$.
A: You could also use the ratio test: $$\left\lvert\frac{a_{n+1}}{a_n}\right\rvert=\left\lvert\frac{4^{n+1}x^{(n+1)^2}}{4^nx^{n^2}}\right\rvert=4\left\lvert x\right\rvert^{2n+1}$$
This ratio converges to $0$ if $\left\lvert x\right\rvert<1$, and diverges to $\infty$ if $\left\lvert x\right\rvert>1$. So the radius of convergence is $1$.
