# Ratio Test and $\sum \frac{n^n}{4^n\cdot n!}$

I'm having trouble showing that the series $\sum_{n=1}^\infty \frac{n^n}{4^n\cdot n!}$ converges.

The only hint I've been given is that the $\lim_{n\to\infty} (1 + 1/n)^n = e$.

I've tried setting it up, and have worked the limit down to:

$$\lim_{n\to\infty}\left(\frac{n}{(n+1)}\right)^2$$

Can anyone give me a hint as to where to go next?

To prove the convergence through the ratio test is straightforward.

As a side note, I just like to mention that by Lagrange's inversion formula we have: $$\forall x\in\mathbb{R}:|x|<\frac{1}{e},\quad W_0(x) = \sum_{n\geq 1}\frac{(-1)^{n-1} n^{n-1}}{n!}\,x^n \tag{1}$$ where $W_0$ is the principal branch of the Lambert W function, i.e. the inverse function of $xe^x$.

$(1)$ gives: $$\sum_{n\geq 1}\frac{n^n}{4^n n!}=\frac{1}{4}\cdot W_0'\left(-\frac{1}{4}\right)=-\frac{W_0\left(-\frac{1}{4}\right)}{1+W_0\left(-\frac{1}{4}\right)} \tag{2}$$ so the value of our series can be found by solving: $$x e^{x} = -\frac{1}{4}\tag{3}$$ through Newton's method, for instance. Since the solution of $(3)$ is between $-\frac{4}{11}$ and $-\frac{5}{14}$,

$$\frac{5}{9}\leq\sum_{n\geq 1}\frac{n^n}{4^n n!}\leq\frac{4}{7}.\tag{4}$$

The limit test involves the limit:

$$\begin{split} L &= \lim_{n \to \infty}\frac{(n+1)^{n+1}}{4^{n+1}(n+1)!}\cdot\frac{4^nn!}{n^n} \\ &= \lim_{n \to \infty}\frac{4^n}{4^{n+1}}\cdot\frac{n!}{(n+1)!}\cdot\frac{(n+1)^{n+1}}{n^n} \\ &= \lim_{n \to \infty}\frac{(n+1)^{n+1}}{4(n+1)n^n} \\ &= \lim_{n \to \infty}\frac14 \frac{(n+1)^n}{n^n} \\ &= \lim_{n \to \infty}\frac14 (1 + \frac1n)^n \end{split}$$

From the last step, you can use your hint to find $L$.

Let: $$a_n = \frac{n^n}{4^n\cdot n!}.$$ Then: $$\frac{a_{n+1}}{a_n} = \frac{1}{4}\cdot\frac{(n+1)^{n+1}}{n^n}\cdot\frac{1}{n+1} = \frac{1}{4}\left(1+\frac{1}{n}\right)^n\to\frac{e}{4}$$ so the ratio test gives that the limit is zero.

Form the ratio $\frac{a_{n+1}}{a_n}$ as

$$\frac{a_{n+1}}{a_n}=\frac{4^nn!(n+1)^{n+1}}{4^{n+1}(n+1)!n^n}=\frac14 \left(1+\frac1 n\right)^n\to \frac14 e<1$$

There seems to be an error in your work. Using the ratio test should lead you down the following path:

$$\lim_{n\to\infty} \frac{\frac{(n+1)^{n+1}}{4^{n+1}(n+1)!}}{\frac{n^n}{4^n n!}}= \lim_{n\to\infty} \frac14 \frac{(n+1)^n}{n^n} \frac{(n+1)n!}{(n+1)!} = \lim_{n\to\infty} \frac14 \left( \frac{n+1}{n}\right)^n = \lim_{n\to\infty} \frac14 \left( 1+ \frac1n \right)^n$$

You can apply the hint your were given to finish the problem.