Radius of convergence for $\sum_{n=0}^\infty n^nx^n$ and $\sum_{n=0}^\infty \frac {(-3)^n}{n}(x+1)^n$

Question

How can one calculate the radius of convergence for the following power series:

$$\sum_{n=0}^\infty n^nx^n$$

and

$$\sum_{n=0}^\infty \frac {(-3)^n}{n}(x+1)^n$$

Regarding the first one I know that for $\sum_{n=0}^\infty x^n$ we get

$$r= \frac{1}{\lim_{n \to \infty} \sqrt[n]{|a_n|} }$$ $$r= \frac{1}{\lim_{n \to \infty} \sqrt[n]{|1|} }$$ $$r=1$$

But how does it work for the $n^n$?

Regarding the second one I tried it out on paper, but don't get anywhere because of the $(x+1)^n$

Answer 1

$ \lim \sup \sqrt[n]{n^n}= \lim \sup n= \infty.$ Hence, the first power series has radius of convergence $=0.$

If $a_n= \frac{(-3)^n}{n}$, then $ \lim \sup \sqrt[n]{|a_n|}=3$, thus $r=1/3.$

Answer 2

$$\lim_{n\to\infty}\left(\dfrac{(n+1)^{n+1}x^{n+1}}{n^n}x^n\right)=\lim_{n\to\infty} n\lim_{n\to\infty}\left(1+\dfrac1n\right)^nx=?$$

For

$$\dfrac{(-3)^n(x+1)^n}n=\dfrac{\left(-3(x+1)\right)^n}n$$

Use ratio test