$\sum_{n=0}^\infty \frac{(1+\frac{1}{n})^n\cdot n!}{n^n}x^n$ prove if it converge or diverge for $x=-e$

Question

$$\sum_{n=0}^\infty \frac{(1+\frac{1}{n})^n\cdot n!}{n^n}x^n$$

Hello, this power series has a radius of $e$, but I cannot conclude if it diverges or converges in $x = -e$, I didn't succeed writing it formally. I would to receive hints or a valid solution, thanks.

Answer 1

Use Stirling's approximation as noted by @Robert Z $$n! \sim \sqrt{2 \pi n} \frac{n^n}{e^n}$$ So $$ \sum_{n=0}^\infty \frac{(1+\frac{1}{n})^n\cdot n!}{n^n}x^n \sim \sqrt{2 \pi}\sum_{n=0}^\infty f(n)$$ where $$f(n) = (1+\frac{1}{n})^n\cdot \sqrt{ n} \frac{1}{e^n}x^n$$ For $x = -e$, we have $$f(n) = (-1)^n(1+\frac{1}{n})^n\cdot \sqrt{ n} $$ Consider Let's work with $\vert f(n) \vert$ $$g(n) = \log \vert f(n) \vert = \underbrace{n \log(1+\frac{1}{n})}_{\rightarrow 1} + \underbrace{\frac{1}{2} \log n}_{\rightarrow \infty} \rightarrow +\infty$$ So $$\vert f(n) \vert \rightarrow + \infty \neq 0$$ Hence, the series diverges.

Answer 2

You can prove that $a_n=\frac{(1+\frac{1}{n})^n\cdot n!}{n^n}e^n$ is a strict increasing sequence. So its limit is not zero and this implies your series is divergent! $$\frac{a_{n+1}}{a_n}=\frac{(1+\frac{1}{n+1})^{n+1}\cdot (n+1)!}{{(n+1)}^{n+1}}e^{n+1}\frac{n^n}{e^n(1+\frac{1}{n})^n\cdot n!} =\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^n}\frac{e}{(1+\frac{1}{n})^n}>1.$$ Here the sequence $\{(1+\frac{1}{n})^n\}$ is strict increasing tending to $e$.

Answer 3

Note that

$$\tag 1\ln \left (\frac{n!e^n}{n^n}\right ) = \sum_{k=1}^{n} \ln k + n - n\ln n.$$

Now $\ln x$ is increasing, and this implies $\sum_{k=1}^{n} \ln k \ge \int_1^n \ln x\, dx.$ That integral equals $n\ln n -n +1.$ Thus the expression in $(1) $ is at least $1.$ Therefore $\dfrac{n!e^n}{n^n}$ is at least $e,$ which implies your series diverges.

Answer 4

$f_m(x) =\sum_{n=1}^{m} \frac{(1+\frac{1}{n})^n\cdot n!}{n^n}x^n $

(The usual)

$\begin{array}\\ (1+1/n)^n &=\exp(n\ln(1+1/n))\\ &=\exp(n(1/n-1/(2n^2)+O(1/n^3)))\\ &=\exp(1-1/(2n)+O(1/n^2))\\ &=e\exp(-1/(2n)+O(1/n^2))\\ &=e(1-1/(2n)+O(1/n^2))\\ \end{array} $

so, since $n! \approx cn^{n+1/2}e^{-n} $,

$\begin{array}\\ f_m(x) &=\sum_{n=1}^{m} \dfrac{(1+\frac{1}{n})^n\cdot n!}{n^n}x^n\\ &=\sum_{n=1}^{m} \dfrac{(e(1-1/(2n)+O(1/n^2)))\cdot n!}{n^n}x^n\\ &=\sum_{n=1}^{m} \dfrac{e n!}{n^n}x^n -\sum_{n=1}^{m} \dfrac{e\cdot n!}{2n\cdot n^n}x^n +O(\sum_{n=1}^{m} \dfrac{ n!}{n^2n^n}x^n)\\ &=\sum_{n=1}^{m} \dfrac{e cn^{n+1/2}e^{-n}}{n^n}x^n -\sum_{n=1}^{m} \dfrac{e\cdot cn^{n+1/2}e^{-n}}{2n\cdot n^n}x^n +O(\sum_{n=1}^{m} \dfrac{ cn^{n+1/2}e^{-n}}{n^2n^n}x^n)\\ &=ec\sum_{n=1}^{m} \sqrt{n}(x/e)^n -(ec/2)\sum_{n=1}^{m} n^{-1/2}(x/e)^n +O(\sum_{n=1}^{m} n^{-3/2}(x/e)^n)\\ \end{array} $

At $x=e$, the third series converges, the second series diverges slowly like $\sqrt{m}$, and the first series diverges like $m^{3/2}$.

So the overall sum diverges.

This holds even if we take more terms in the Stirling approximation, since the first sum remains the same and only the second sum is changed.

Answer 5

A necessary condition for a series $\sum _{k=0} ^{\infty} a_k $ to converge is that $\lim _{k \rightarrow \infty} a_k =0 $. You can prove this condition does not hold for $x=-e$, hence the power series diverges at $x=-e$. In fact, using Stirling approximation $a_k \approx e\sqrt{2\pi k} \left( \frac{x} {e} \right) ^k$ for $k \rightarrow \infty$.