What I did wrong in finding the radius of convergence for this problem?

Question

I'm asked to find the radius of convergence for summation from n = 1 to $\infty$ of $\sum n!x^n/(4n^n)$. I used the ratio test because of the factorial, and I eventually got a whole bunch of cancellations that just left me with $|x|$, and so if $|x| < 1$, $R=1$, but this answer is apparently wrong. I redid the problem a couple times very slowly, but I kept getting the same answer. Did I make a mistake somewhere, or am I using the wrong approach?

Edit: here is my approach

$$\lim\left|\frac{(n+1)n!x^nx}{4(n+1)^n(n+1)} \frac{4n^n}{n!x^n}\right|$$

The (n+1) terms cancel, the n! terms cancel, the 4's cancel, the n^n's cancel, the x^n terms cancel, leaving |x|

Answer 1

It's actually not hard to show that $(n+1)^n\geq 2n^n$ for $n\geq 1$.

The first two terms of $(n+1)^k$ are $n^k + k\cdot n^{k-1}$. So $(n+1)^k\geq n^k+k\cdot n^{k-1}$ when $n\geq 0$ and $k\geq 1$.

When $k=n$, then $(n+1)^n\geq n^n+n\cdot n^{n-1}=2n^n$.

This in turn means that $\left(\frac{n}{n+1}\right)^n\leq \frac{1}{2}$ for all $n\geq 1$. In particular, the limit cannot be $1$, so you cannot "cancel" $n^n$ and $(n+1)^n$.

In fact, $$\left(\frac{n}{n+1}\right)^n\to e^{-1}.$$

Answer 2

I'm not sure what you did where you wrote "here is my approach" (though I think there are several mistakes...), but it seems to be pretty direct:

$$a_n:=\frac{n!}{4n^n}\implies\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{4(n+1)^{n+1}}\cdot\frac{4n^n}{n!}=\frac{n^n}{(n+1)^n}=\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e$$

and thus the convergence radis is $\;R=e\;$ and we have convergence for

$$\frac{|x|}e<1\iff |x|<e\iff -e<x<e$$