I need to find the radius of the convergence of $\sum_{n=1}^{\infty}3^{n}z^{n^{2}}$ using the Cauchy-Hadamard formula. I'm not feeling 100% proficient at this method, however, so I'm asking 1) if what I did below is correct, and 2) if not, what is the correct way to do it/correct answer?
Here's what I did:
Let $a_{n}=3^{k}$. Then, $\limsup_{n\to \infty}\sqrt[n]{|3^{k}|} = \limsup_{n \to \infty}\sqrt[k^{2}]{|3|^{k}}=\limsup_{n \to \infty} (3^{k})^{1/k^{2}} = \limsup_{n \to \infty}3^{1/k} = \limsup_{n \to \infty}3^{1/\sqrt{n}} = 3^{0} = 1 = \Lambda$,
And, $R = \frac{1}{\Lambda} = 1$ is the radius of convergence.
I also did it using the Ratio Test, letting $u_{n}=3^{n}z^{n^{2}}$ and $u_{n+1} = 3^{n+1}z^{(n+1)^{2}} = 3^{n+1}z^{n^{2}+2n+1}$. Then, $\displaystyle \lim_{n \to \infty} \left\vert \frac{u_{n+1}}{u_{n}}\right\vert = \lim_{n \to \infty}\left\vert \frac{3^{n+1}z^{n^{2}+2n+1}}{3^{n}z^{n^{2}}}\right\vert = \lim_{n \to \infty}\left\vert 3\cdot z^{2n+1} \right\vert = \lim_{n \to \infty}3 |z|^{2n+1}$, which converges only if $|z|<1$, so $R = 1$.
But, again, not sure if I did this right. Could use a second pair of eyes while learning this stuff. Thanks.
Best Answer
Another approach: If $|z|>1,$ the series diverges simply because the terms don't approach $0.$ If $|z|<1,$ then $|z|^n \to 0,$ hence $|z|^n<1/4$ for large $n.$ Hence the $n$th term of the series in absolute value is $(3|z|^n)^n < (3/4)^n$ for large $n.$ That implies convergence of the series by the comparison test. The radius of is therefore $1.$