Radius and domain of convergence for $\sum _{n=1}^{\infty}2^n x^{n^2}$ Another Question on Radius of convergence :
Calculate Radius and domain of convergence for $$\sum _{n=1}^{\infty}2^n x^{n^2}$$
I used the formula $\lim_{n\rightarrow \infty} |\frac{a_n}{a_{n+1}}|$ to calculate radius of convergence.
I have $\lim_{n\rightarrow \infty} |\frac{a_n}{a_{n+1}}|=\lim_{n\rightarrow \infty} |\frac{2^n}{2^{n+1}}|=\frac{1}{2}$ 
So,radius of convergence is $\frac{1}{2}$ for $\sum _{n=1}^{\infty}2^n x^n$ but how would i use this to calculate radius of convergence for $$\sum _{n=1}^{\infty}2^n x^{n^2}$$
If i have series something like $\sum _{n=1}^{\infty}2^n (x^n)^2$
I would have just write radius of convergence is $\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$
But in this case i do not know what to do.. please help me to see this.
Another point is I some how tried to use : For $\sum_{n=1}^{\infty}a_nx^n$ radius would be $R$ where
$$\frac{1}{R}=\lim \sup \sqrt[n]{|a_n|}$$
My book says that if I have some thing like  $$\sum _{n=1}^{\infty}a_n x^{n^2}$$ I could use same formula with $$\frac{1}{R}=\lim \sup \sqrt[n^2]{|a_n|}$$
I am not very sure if i can really use it but if i use that, I would get :
$$\frac{1}{R}=\lim \sup \sqrt[n^2]{2^n}=\sqrt{2}$$
SO, radius of convergence would be $\frac{1}{\sqrt{2}}$
which is not much different from what i have got with previous case of $\sum _{n=1}^{\infty}2^n (x^n)^2$ 
I would like to know more about something similar to this..
Even if i have $\sum _{n=1}^{\infty}a_n x^{n^2}$ or $\sum _{n=1}^{\infty}a_n x^n$, the radius of convergenc is not changing.
Please help me to look into this in more detail.
Thank you 
 A: Since your series has the form
$$\sum_{n}a_n x^{\color{red}{n^2}}$$
and not 
$$\sum_{n}a_n x^{\color{green}{n}}$$
hence we use the ratio test in this way: let $u_n=a_n x^{n^2}$
$$\lim_{n\to\infty}\left|\frac{u_{n+1}}{u_n}\right|=\lim_{n\to\infty}2|x|^{2n+1}<1\iff |x|<1 $$
hence the radius is $R=1$. Do the same thing for using the Cauchy test.
A: Why not use this, which I guess you mentioned in the question?
You'll have 
$$C=\lim\sup\sqrt[n]{|2^n\cdot(x^n)^n|}=\lim\sup(2\cdot|x|^n).$$
If $|x|\lt1$, then $C=0\lt 1$. If $|x|\ge1$, then $C\gt1$. 
A: Define the sequence
$$
a_n=\left\{\begin{matrix} 2^{\sqrt{n}} &\text{if $n=m^2$ for some $m\in\mathbb{N}$}\\
0 &\text{otherwise}
\end{matrix}
\right.
$$
The series in the OP can be expressed as
$$\sum_na_nx^n$$
The radius of convergence is $\frac{1}{L}$ where $L=\limsup_n\sqrt[n]{|a_n|}$ which in this case is
$$L=\lim_{m\rightarrow\infty}\big(2^m)^{1/m^2}=\lim_{m\rightarrow\infty}2^{1/m}=1$$
Thus the series converges for all $x$ such that $|x|<1$.
A: The radius of convergence is $1,$ since obviously the series diverges for $|x|>1,$ and for $|x|<1,$ there exists an $n(x)$ for which $x^{n(x)} < \frac13,$ so for $n>n(x),$ we have
$$ 2^n x^{n^2} < 2^n x^{n(x) n} < (2/3)^n.$$
