Since we know that
given $\sum_{n=0}^{\infty }C_nz^n$, if $\lim_{n\rightarrow \infty }|C_n|^{1/n}$ exists then $R^{-1}=\lim_{n\rightarrow \infty }|C_n|^{1/n}$ where $R$ is the radius of convergence.
But for the power series $\sum_{n=0}^{\infty }2^nz^{n^2}$, when I find the radius of convergence of it, should I find
$\lim_{n\rightarrow \infty }|C_n|^{1/n^2}$? Then $\lim_{n\rightarrow \infty }|C_n|^{1/n^2}=\lim_{n\rightarrow \infty }|2^n|^{1/n^2}$ which is $1$. So the convergence radius is $1$.
Am I correct? If no, how to find the convergence radius of the power series that is not in the form of $\sum_{n=0}^{\infty }C_nz^n$, but like $\sum_{n=0}^{\infty }C_nz^{n^2}$ or $\sum_{n=0}^{\infty }C_nz^{n!}$?
Thanks in advance!
$\endgroup$ 12 Answers
$\begingroup$The Ratio Test works too: \begin{align*} \lim_{n\rightarrow\infty}\left|\frac{2^{n+1}z^{(n+1)^2}}{2^nz^{n^2}}\right| &= 2\lim_{n\rightarrow\infty}|z|^{n^2+2n+1-n^2} \\ &= 2\lim_{n\rightarrow\infty}|z|^{2n+1} \\ &= \begin{cases} \infty & |z|>1 \\ 2 & |z|=1 \\ 0 & |z|<1 \end{cases} \end{align*} By the Ratio Test, $\displaystyle\sum_{n=0}^\infty 2^nz^{n^2}$ converges for $|z|<1$ and diverges for $|z|\geq 1$. In particular, $R=1$.
$\endgroup$ $\begingroup$Using the $n$th root test, $$|2^nz^{n^2}|^{1/n}=2|z|^n\ .$$ If $|z|<1$ this tends to $0$ as $n\to\infty$ so the series converges; if $|z|>1$ this tends to $\infty$ as $n\to\infty$ so the series diverges. Therefore the radius of convergence is $1$.
You could also do it by the ratio test.
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