Compute this radius of convergence

Question

Someone can help me with this radius of convergence please?

$$ \sum_{n=0}^{\infty}{n^{1/n}x^{n}} $$

I tried

$$ r = \lim_{n \to \infty } \frac{n^{\frac{1}{n}}x^{n}}{(n+1)^{\frac{1}{n+1}}(x^{n+1})} $$

After to operate, I can't get to answer

Answer 1

Hint: try using the root test. Note that for $n>1$, we have $$ 1\leq n^{1/{n^2}}\leq n^{1/n} $$

Answer 2

If $\lim \frac{a_n}{a_{n+1}}$ exists, then the radius of convergence is that limit. Since $n^{\frac{1}{n}}\to 1$ and $(n+1)^{\frac{1}{n+1}}\to 1$ , the radius of convergence is 1.

Answer 3

The radius of convergence R was defined to be $$\frac{1}{R} = limsup_{n \to \infty} \sqrt[n]{|a_n|}$$ IF $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n}$$ exists, then this is also $\frac{1}{R}$ but in general the first works better.

In this case $\sqrt[n]{|a_n|} = n^{1/{n^2}} \to 1$ as $n\to \infty$, thus $R = 1$.