# 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?

$$\sum_{n=1}^\infty a_n x^n\;\;,\;\;a_n=\begin{cases}0,&n\text{ is not a square}\\{}\\2^{2k},&n=k^2\end{cases}$$

$$\frac1R=\limsup_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\left(2^{2\sqrt n}\right)^{1/n}=\lim_{n\to\infty}4^{1/\sqrt n}=1$$

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.

• Happy Holidays my friend! Actually, the series is a power series with all coefficients on non-squared terms equal to $0$. Your analysis quite aptly shows that this is true. – Mark Viola Dec 30 '17 at 21:27
• @MarkViola Don't agree: it is not a power series as given formally and thus Cauchy-Hadamard and other tests cannot be applied to the series as given. One must take the coefficients as shown above in order to be able to apply these tests and formulas. And Happy New Year to you and everyody else! – DonAntonio Dec 30 '17 at 21:32

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 $|x|<1$.

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 $|x| < 1$.

• It appears that we posted our solutions at nearly the same time. It is of interest to note that we can also view the series of interest as a power series where the coefficients are zero on all of the non-squared terms. – Mark Viola Dec 30 '17 at 21:09

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 $1$.