Convergence radius of $\sum_{n=1}^\infty (-1)^nn^nx^{n^2} $

Question

I need to find the convergence radius of the series $$\sum_{n=1}^\infty (-1)^nn^nx^{n^2} $$ I have tried using the ratio test $$ \lim\limits_{n \to \infty} \lvert \frac{a_{n+1}}{a_{n}}\rvert $$ and i am left with $$ \lim\limits_{n \to \infty} \lvert nx^{2n} \lvert $$ The exponent on the x is bothering me since i need to bring it outside of the limit in order to check for which values of x will the term be less than 1.

How can I proceed? Thanks

Answer 1

Root test may save you:

$$ \lim_{n\to\infty} |n^n x^{n^2}|^{1/n} = \lim_{n\to\infty} n |x|^n = \begin{cases} \infty, & |x| \geq 1 \\ 0, & |x| < 1. \end{cases} $$

Therefore the series converges if and only if $|x| < 1$ and hence the radius of convergence is 1.

If you want to apply the Cauchy-Hadamard theorem, notice that the coefficients of the power series is given by

$$ \sum_{k=0}^{\infty} (-1)^k k^k x^{k^2} = \sum_{n=0}^{\infty} a_n x^n, \qquad \text{where} \quad a_n = \begin{cases} (-1)^k k^k, & n = k^2 \\ 0, & \text{otherwise}. \end{cases} $$

Consequently

$$ \limsup_{n\to\infty} \sqrt[n]{\smash[b]{|a_n|}} = \lim_{k\to\infty} (k^k)^{1/k^2} = \lim_{k\to\infty} k^{1/k} = 1. $$

This again proves that the radius of convergence is 1.