$\sum_{n=1}^\infty \frac{n!e^n}{n^{n+ \frac{3}{2}}}$ - any ideas for a simple proof of divergence?

Question

I am looking for a simple proof of divergence for the series: $\sum_{n=1}^\infty \frac{n!e^n}{n^{n+\frac{3}{2}}}$

That's a part of the more general problem:

For what values of X is the series $\sum_{n=1}^\infty \frac{n!e^n}{n^{n+X}}$ convergent and for what values is it divergent?

I am not allowed to use Stirling's approximation in the proof.

I've already managed to prove convergence for $X>\frac{3}{2}$ and divergence for $X<\frac{3}{2}$. And now I am stuck with $X=\frac{3}{2}$ - I know the series is divergent (from Stirling approximation and WolframAlpha) but I have no idea for an elementary proof.

This question is related to my previous one about the elementary proof for the simpler case i.e.: $\sum_{n=1}^\infty \frac{n!e^n}{n^n}$ so you might be interested in checking it out: Divergent infinite series $n!e^n/n^n$ - simpler proof of divergence?

Answer 1

$$ n!e^n = \int_{0}^{+\infty} x^n e^{n-x}\,dx \geq \int_{n}^{+\infty} x^n e^{n-x}\,dx =\int_{0}^{+\infty}(n+x)^n e^{-x}\,dx=\\= n^{n+1}\int_{0}^{+\infty}\left[(1+x)e^{-x}\right]^n\,dx$$ so if we manage to prove that $$ \sum_{n\geq 1}\frac{1}{\sqrt{n}}\int_{0}^{+\infty}\left[(1+x)e^{-x}\right]^n\,dx $$ is unbounded we are done. We may recall that $\frac{1}{4^n}\binom{2n}{n}\leq\frac{1}{\sqrt{\pi n}}$ by Wallis product, hence the previous series is bounded below by $$\sqrt{\pi}\sum_{n\geq 1}\int_{0}^{+\infty}\frac{1}{4^n}\binom{2n}{n}\left[(1+x)e^{-x}\right]^n\,dx. $$ Since $$ \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}z^n = \frac{1}{\sqrt{1-z}} $$ for any $z\in(-1,1)$, the divergence of the given series is reduced to the divergence of $$ \int_{0}^{+\infty}\left[\frac{1}{\sqrt{(e^x-1-x)e^{-x}}}-1\right]\,dx$$ which is fairly straightforward by considering the Laurent expansion at the origin of the integrand function - $0$ is a simple pole with residue $\sqrt{2}$.

^{Truth to be told, we have just proved a weak version of Stirling's inequality.}

Answer 2

On similar lines as demonstrated by Jack D'Aurizio it is possible to prove by even more "elementary" means: $$ n!>\left(\frac n e \right)^n\sqrt{\frac{n\pi}2}\tag{1}, $$ which suffices for the proof of divergence by comparison with harmonic series as $$ \frac{n! e^n}{n^{n+\frac32}}>\sqrt{\frac{\pi}2}\frac1n. $$

Proof of (1): $$ \begin{array}{ll} n!&=\int\limits_0^\infty x^ne^{-x}dx\stackrel{x\mapsto n+\sqrt n t} =\int\limits_{-\sqrt n}^\infty (n+\sqrt n t)^ne^{-n-\sqrt n t}\sqrt n dt\\ &=\left(\frac n e \right)^n\sqrt{n}\int\limits_{-\sqrt n}^\infty \left(1+\frac t {\sqrt n}\right)^ne^{-\sqrt n t}dt >\left(\frac n e \right)^n\sqrt{n}\int\limits_{0}^{\infty} \left(1+\frac t {\sqrt n}\right)^ne^{-\sqrt n t}dt\\ &=\left(\frac n e \right)^n\sqrt{n}\int\limits_{0}^{\infty} e^{n\log\left(1+\frac t {\sqrt n}\right)} e^{-\sqrt n t}dt >\left(\frac n e \right)^n\sqrt{n}\int\limits_{0}^{\infty} e^{n\left[\frac t {\sqrt n}-\frac12\left(\frac t {\sqrt n}\right)^2 \right]} e^{-\sqrt n t}dt\\ &=\left(\frac n e \right)^n\sqrt{n}\int\limits_{0}^{\infty} e^{-\frac{t^2}2} dt =\left(\frac n e \right)^n\sqrt\frac{n\pi}2. \end{array} $$

Note that the RHS of (1) is two times less than the Stirling expression. The reason for the underestimation is neglected "negative $t$" part of the integral, which (as well as the "positive" part) tends to $\sqrt\frac{\pi}2$ with increasing $n$.

Answer 3

Let $a_n=n\cdot\frac{n!e^n}{n^{n+3/2}}$. We shall prove $a_n$ has a nonzero limit, so by comparison to harmonic series the series in question diverges.

It's easy to compute $$\frac{a_n}{a_{n+1}}=\frac{1}{e}\left(1+\frac{1}{n}\right)^{n+1/2}.$$ Taking logarithm and using the Taylor expansion, we have $$\log{a_n}-\log{a_{n+1}}=-1+\left(n+\frac{1}{2}\right)\log\left(1+\frac{1}{n}\right)=-1+\left(n+\frac{1}{2}\right)\left(\frac{1}{n}-\frac{1}{2n^2}+O\left(\frac{1}{n^3}\right)\right)\\ =-1+1+\frac{1}{2n}-\frac{1}{2n}-\frac{1}{4n^2}+O\left(\frac{1}{n^2}\right)=O\left(\frac{1}{n^2}\right),$$ which means the series $$\sum_{n=1}^\infty(\log{a_n}-\log{a_{n+1}})$$ converges. Now we observe the series is telescoping, so that $$\sum_{n=1}^k(\log{a_n}-\log{a_{n+1}})=\log a_1-\log a_{k+1},$$ so $\log a_n$ has a limit, hence $a_n$ has a nonzero limit.

Answer 4

To start with, a straightforward adaptation of the proof of the integral test will show that $$\log(n!) = \sum_{k=1}^n \log(n) > \int_1^{n-1} \log x\,dx = (n-1) \log(n-1) - (n-1).$$ Similarly, $\log(n!) < n \log n - n$; so $\log(n!)$ is asymptotic to $n \log n$ as $n \to \infty$.

Now, after that first bootstrapping step, let us set $a_n := \log(n!) - n \log n$ and investigate the asymptotic behavior of $a_n$. We see that $$\Delta a_n = a_{n+1} - a_n = \log(n+1) - (n+1) \log(n+1) + n \log(n) = \\ -n \log\left(1 + \frac{1}{n}\right) = -1 + \frac{1}{2n} - O(1/n^2).$$ Therefore, $a_n \sim -n$ as $n \to \infty$, and furthermore $a_n + n \sim \frac{1}{2} \log n$ as $n \to \infty$.

Now, setting $b_n := \log(n!) - (n \log n - n + \frac{1}{2} \log n)$, we will similarly be able to show that $\Delta b_n = O(1/n^2)$, so $b_n$ is a convergent sequence. Based on this, we can conclude that $$n! = \Theta(n^{n+1/2} e^{-n}) \mathrm{~as~}n \to \infty$$ which is sufficient to show that the original series is divergent.

Answer 5

Let $$ a_n=\frac{e^nn!}{n^{n+3/2}} $$ It suffices to show that $$ \frac{a_{n+1}}{a_n}\ge 1-\frac{1}{n}=\frac{n-1}{n} $$ since this implies in turn that $$ a_n=a_2\cdot\frac{a_3}{a_2}\frac{a_4}{a_3}\cdots \frac{a_n}{a_{n-1}}\ge a_2\cdot\frac{2}{3}\cdot\frac{3}{4}\cdots\frac{n-1}{n}=\frac{2a_2}{n} $$ But $$ \frac{a_{n+1}}{a_n}=\frac{e}{\left(1+\frac{1}{n}\right)^{n+3/2}} $$ So it suffices to show that $$ \frac{e}{\left(1+\frac{1}{n}\right)^{n+3/2}}\ge 1-\frac{1}{n} $$ or that $$ e\ge (1-x)(1+x)^{1/x+3/2} \qquad \text{for all $x\in(0,1)$} $$ or that $$ 1\ge \log (1-x)+\left(\frac32+\frac1x\right)\log(1+x)\qquad \text{for all $x\in(0,1)$}. $$ But $$ \log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots, \qquad \log(1-x)=x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots, $$ and hence $$ \log (1-x)+\left(\frac32+\frac1x\right)\log(1+x)=\cdots=1+\sum_{n=2}^\infty\left(-\frac{1}{n}+\frac{3(-1)^{n+1}}{2n}+\frac{(-1)^n}{n+1}\right)x^n\le 1, $$ since $$ -\frac{1}{n}+\frac{3(-1)^{n+1}}{2n}+\frac{(-1)^n}{n+1}=-\frac{1}{2n(n+1)}\left(n(2+(-1)^n)+2+3(-1)^n\right)\le -\frac{n-1}{2n(n+1)}\le 0. $$