Finding the convergence interval of $\sum_{n=0}^\infty\frac{n!x^n}{n^n}$.

Question

I want to find the convergence interval of the infinite series $\sum\limits_{n=0}^\infty \dfrac{n!x^n}{n^n}$.

I will use the ratio test: if I call $u_n = \dfrac{n!x^n}{n^n}$, the ratio test says that, if the following is true for some values of $x$, the series will be convergent for these values of $x$:

$$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|<1$$

So, I will first calculate the value of $\left|\dfrac{u_{n+1}}{u_n}\right|$:

$$\left|\dfrac{\dfrac{(n+1)!x^{n+1}}{(n+1)^{n+1}}}{\dfrac{n!x^n}{n^n}}\right|=\dfrac{(n+1)!|x|^{n+1}}{(n+1)^{n+1}}\times\dfrac{n^n}{n!|x|^n}=\frac{(n+1)n^n|x|}{(n+1)^{n+1}}=|x|\left(\frac{n}{n+1}\right)^n$$

So, $\lim\limits_{n\to+\infty}\left|\dfrac{u_{n+1}}{u_n}\right|$ becomes:

$$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|=\lim_{n\to+\infty}|x|\left(\frac{n}{n+1}\right)^n=|x|\lim_{n\to+\infty}\left(\frac{n}{n+1}\right)^n$$

Now I must evaluate the value of $\lim\limits_{n\to+\infty}\left(\dfrac{n}{n+1}\right)^n$. For this, let $y = \left(\dfrac{n}{n+1}\right)^n$; so, instead of calculating $\lim\limits_{n\to+\infty}y$, I will first calculate $\lim\limits_{n\to+\infty}\ln y$:

$$\lim_{n\to+\infty}\ln y=\lim_{n\to+\infty}\ln \left(\dfrac{n}{n+1}\right)^n=\lim_{n\to+\infty}n\ln\left(\frac{n}{n+1}\right) =\lim_{n\to+\infty}\frac{\ln\left(\frac{n}{n+1}\right)}{\frac{1}{n}}$$

Applying L'Hôpital's rule:

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

Now, since we know that $\lim\limits_{n\to+\infty}\ln y = -1$, we have that:

$$\lim_{n\to+\infty}y=\lim_{n\to+\infty}e^{\ln y} = e^{-1} = \frac{1}{e}$$.

Substituting this back into the expression $\lim\limits_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right| = |x|\lim\limits_{n\to+\infty}\left(\frac{n}{n+1}\right)^n$, we have that the limit of $\left|\dfrac{u_{n+1}}{u_n}\right|$ as $n\to+\infty$ is:

$$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|=\frac{|x|}{e}$$

Therefore, the series will certainly be convergent for the values of $x$ for which $\dfrac{|x|}{e}<1$, that is, $|x|<e$.

So, I know that the series is convergent for $-e < x < e$, but I have to test whether the series is convergent at $x = e$ or $x = -e$. That is, I have to test whether $\sum\limits_{n=0}^{\infty} \dfrac{n!e^n}{n^n}$ and $\sum\limits_{n=0}^{\infty} \dfrac{(-1)^nn!e^n}{n^n}$ are convergent. Since these limits don't approach zero, I know they are both divergent, but I'm not sure how to find the limits, because of the factorial function. Also, I can't use integral test here, because of the factorial. Probably I should use comparison test, but I haven't found any divergent series to which to compare it.

Any hints?

Thank you in advance.

---

Edit: Using the suggestion by Ragib Zaman in the answer below, since the Taylor polynomial $P_n(x)$ of $e^x$ at $a=0$ is $$e^x = 1 + x + \dfrac{x^2}{2!} + \cdots + \dfrac{x^n}{n!}+\cdots,$$ if we substitute $n$ for $x$ we see that $e^n>\dfrac{n^n}{n!}$; therefore, $\dfrac{n!e^n}{n^n} > 1$, and, thus, we show that $\sum\limits_{n=0}^{\infty} \dfrac{n!e^n}{n^n}$ is divergent, because its term doesn't approach zero.

$\sum\limits_{n=0}^{\infty} \dfrac{(-1)^nn!e^n}{n^n}$ is also divergent, because, although the absolute value of the ratio between two successive terms, $|e|\left(\frac{n}{n+1}\right)^n$, approaches 1 as $n\to\infty$, it approaches 1 from values bigger than 1; therefore, the absolute value of a term is always greater than the absolute value of the previous term.

Answer 1

Since $\displaystyle e^x > \frac{x^n}{n!}$ for all $x>0, n \in \mathbb{N}$, letting $x=n$ gives $\displaystyle \frac{n!e^n}{n^n}>1$ so the terms don't even approach zero.

Answer 2

This may be overkill but we have $$n!\approx n^{n+1/2}e^{-n}$$ by Stiriling's formula when $n$ goes very large. Thus the original sum becomes approximately $$n^{1/2}\frac{x^{n}}{e^{n}}$$ So if $x\ge e$ this series obviously will not converge. When $0<x<e$ name $\frac{x}{e}=k<0$, then we have $$\sum n^{1/2}k^{n}$$ the ratio test give $$(\frac{n+1}{n})^{1/2}k$$ which becomes less than 1 when $n$ goes very large.

The case that $x$ is negative is similar by the alternating series test.