Convergence of the power series $\sum_{n=1}^\infty\frac{(1+n)^n}{n!}x^n $

Question

Consider the power series $$\sum_{n=1}^\infty\frac{(1+n)^n}{n!}x^n $$ Now the ratio test shows that the given power series converges absolutely $|x|< 1/e$ and diverges for $|x|>1/e$. But all the tests I know(Raabe's test) are failing at $|x|=1/e$. Any help will be highly apprecieated.

Answer 1

For $x=\frac 1 e$ we have

$$\frac{(1+n)^n}{n!e^n}$$

and using Stirling's approximation $n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$ we have

$$\frac{(1+n)^n}{e^nn!}\sim\frac1{\sqrt{2 \pi n}}\left(1+\frac1n\right)^n$$

then recall that by Bernoulli

$$\left(1+\frac1n\right)^n\ge 2$$

Answer 2

Call $a_n=\frac{(1+n)^n}{n!}x^n$. If you use Raabe's test for $x=\frac1e$, you obtain $$\frac{(1+n)^n(n+1)!e}{(2+n)^{n+1}n!}n-n=en\left(1+\frac1{n+1}\right)^{-n-1}-n=$$

Substitute $t=\frac1{n+1}$ $$=\left(\frac{e}{t}-e\right)(1+t)^{-1/t}-\frac1t+1=\frac{e(1+t)^{-1/t}-1}{t}+1-e(1+t)^{-1/t}$$

And you should recognize, among the terms up there, the incremental ratio at $0$ of the continuous function $$f(t)=\begin{cases}e(1+t)^{-1/t}&\text{if }t>-1\land t\ne 0\\ 1&\text{if }t=0\end{cases}$$ which has derivative $$f'(t)=\begin{cases}e\left(\frac{\ln(1+t)-t}{t^2}+\frac{\ln(1+t)}t\right)(1+t)^{-\frac1t-1}&\text{if }t>-1\land t\ne 0\\ \frac 1{2}&\text{if }t=0\end{cases}$$

So $$\lim_{n\to \infty} n\left(\frac{a_n}{a_{n+1}}-1\right)= \lim_{t\to 0}\frac{e(1+t)^{-1/t}-1}{t}+1-e(1+t)^{-1/t}=f'(0)+1-f(0)=\frac12$$

And so the series is divergent for $x=\frac1e$.

The previous estimate on $n\frac{a_n(1/e)}{a_{n+1}(1/e)}-n=n\left\lvert \frac{a_n(-1/e)}{a_{n+1}(-1/e)}\right\rvert-n$ implies that eventually $$\left\lvert \frac{a_n(-1/e)}{a_{n+1}(-1/e)}\right\rvert\ge 1+\frac1{4n}>1$$

and thus that $\lvert a_{n}(-1/e)\rvert> \lvert a_{n+1}(-1/e)\rvert$ eventually. Therefore, $\sum_{k=0}^\infty a_n(-1/e)$ converges by Leibniz.