Test the convergence of the series given below

Question

Test the convergence of the series for $x>0$ $$1+\frac{1}{2}x+\frac{2!}{3^2}x^2+\frac{3!}{4^3}x^3+\dots$$

Attempt: Let $u_n=\frac{n!}{(n+1)^n}x^n$ then $$\lim_{n\to\infty} \frac{u_{n+1}}{u_n}=\lim_{n\to\infty}\frac{x}{\left(\frac{n+2}{n+1}\right)^{n+1}}=x/e.$$ Then by ratio test of convergence of series,

(1) if $x<e$, $\sum u_n$ is convergent,

(2) if $x>e$, $\sum u_n$ is divergent,

(3) if $x=e$, ratio test fails

Now we have to test the convergence of the following series for the 3rd case ($x=e$)

$$\sum v_n=\sum\frac{n!}{(n+1)^n}e^n$$.

I have applied Cauchy's root test but it fails. Please solve the remaining part of the problem by using a suitable easy test.

Answer 1

Stirling's formula gives $n! \geq \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$. It follows $$\sum\limits_{n = 0}^{\infty} \frac{n!}{(n+1)^n}e^n \geq \sum\limits_{n = 0}^{\infty} \sqrt{2\pi n} \left(\frac{n}{n+1}\right)^n.$$ As the summand diverges, the series diverges.

EDIT: Instead of Stirling, you could also use the simpler estimate $$\log(n!) = \sum\limits_{k = 1}^n \log(k) \geq \int_1^n \log(x) \, \mathrm{d}x = n\log(n) - n +1,$$ which gives $n! \geq n^ne^{-n+1}$. This also leads to a series that obviously diverges.

Answer 2

Hint: Use Stirling's approximation for the factorial for large $n$:

$$n! \approx \sqrt{2\pi n} \bigg(\frac{n}{e}\bigg)^n$$

to view the long term behavior of your summand.

Answer 3

By Stirling's approximation \begin{align} n!&=\sqrt{2\pi n}n^n\mathrm{e}^{-n}\left(1+\mathcal{O}_{n\to+\infty}\left(\frac{1}{n}\right)\right) \end{align} we find \begin{align} \frac{n!}{(n+1)^n}\mathrm{e}^{n}&=\sqrt{2\pi n}\left(\frac{n}{n+1}\right)^n\left(1+\mathcal{O}_{n\to+\infty}\left(\frac{1}{n}\right)\right)=\sqrt{2\pi n}\mathrm{e}^{-1}\left(1+\mathcal{O}_{n\to+\infty}\left(\frac{1}{n}\right)\right) \end{align} which does not tends to 0 as $n\to+\infty$,whence the divergence.