How to determine the convergence of the series $\sum_{n=1}^\infty\frac{e^nn!}{n^n}$?

Question

$$\sum_{n=1}^\infty\frac{e^nn!}{n^n}$$ we cant use the ratio test nor the root test, because the limit will be 1. the main question was to determine whether or not the series $$\sum_{n=1}^\infty\frac{a^nn!}{n^n}$$ converge, for every $a\in\mathbb R$. I found that for every $-e < a < e$ the series absoulutly converges. For every $a>e$ or $a<-e$ the series does not converge (also not conditional convergence) so now I have to determine if for $a=e$ or $a=-e$ the series will converge. (hopefully it will converge for $a=e$ and therefor also for $a=-e$. other wise i'll have to find out also for $a=-e$) thanks. by the way, im not familier with tools like Rabbe's test or Sterling's formula, so solutions using other ideas will be great. thanks

Answer 1

The simple bound $n!>(n/e)^ne$, which can be found by approximating the logarithm of the factorial with an integral, implies $\frac{e^nn!}{n^n}>e$. So the terms in $\sum_{n=1}^\infty\frac{e^nn!}{n^n}$ don't even tend to $0$, and the series diverges. The same is true of $\sum_{n=1}^\infty\frac{(-e)^nn!}{n^n}$.

Answer 2

Without Stirling's formula
When the ratio test is inconclusive, we can try Gauss's test. (An extension of the ratio test.)

Let $u_n > 0$ and suppose $$ \frac{u_n}{u_{n+1}} = 1 + \frac{h}{n} + O(n^{-r}), \qquad h \in \mathbb R,r>1 $$ If $h>1$, then $\sum u_n$ converges; if $h \le 1$ then $\sum u_n$ diverges.

In our case, $$ u_n := \frac{e^n n!}{n^n}, \\ \frac{u_n}{u_{n+1}}= \frac{1}{e}\left(\frac{n+1}{n}\right)^n = 1-\frac{1}{2n} +O(n^{-2}) $$ So we apply Gauss's test with $h=-1/2, r=2$ and conclude the series diverges.

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Explanation for the approimation. $$ \lim_{n\to \infty} \frac{1}{e}\left(\frac{n+1}{n}\right)^n =\frac{1}{e}\;\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = \frac{e}{e} = 1 $$ so the first term is $1$. Next I claim $$ \left(\frac{1}{e}\left(\frac{n+1}{n}\right)^n - 1\right)n \to -\frac{1}{2} $$ so the next term is $-\frac{1}{2n}$. For that computation: $$ \left(\frac{1}{e}\left(\frac{n+1}{n}\right)^n - 1\right)n = \frac{(1+\frac{1}{n})^n-e}{e/n} $$ This is indeterminate of the form $0/0$. L'Hopital suggests this limit is the same as the limit of $$ \frac{(1+\frac{1}{n})^n\left[\log(1+\frac{1}{n})-\frac{1}{n(1+1/n)}\right]}{-\frac{e}{n^2}} = -\frac{(1+\frac{1}{n})^n}{e}\;\cdot\; \left[n^2\log\left(1+\frac{1}{n}\right)-\frac{n^2}{n+1}\right]; $$ Now $-\frac{(1+\frac{1}{n})^n}{e} \to -1$, and to compute the other factor: \begin{align*} \log\left(1+\frac{1}{n}\right) &= \frac{1}{n} - \frac{1}{2n^2} + O(n^{-3}) \\ n^2\log\left(1+\frac{1}{n}\right) &= n - \frac{1}{2} + O(n^{-1}) \\ \frac{n^2}{n+1} &= n - 1 + O(n^{-1}) \\ n^2\log\left(1+\frac{1}{n}\right)- \frac{n^2}{n+1} &= \frac{1}{2} + O(n^{-1}) \\ n^2\log\left(1+\frac{1}{n}\right)- \frac{n^2}{n+1} &\to \frac{1}{2} \end{align*} so that $$ -\frac{(1+\frac{1}{n})^n}{e}\;\cdot\; \left[n^2\log\left(1+\frac{1}{n}\right)-\frac{n^2}{n+1}\right] \to -\frac{1}{2} $$ as claimed.

Answer 3

So, there's much simpler solution wihtout using any advanced tools: prove with induction that $ \frac{e^n \cdot n!}{n^n} \geq 1 $ for all $ n \in N $. for n=1 its trivial. assume for $ n\in N $ so : $ \frac{e^{n}\cdot e\cdot n!\cdot\left(n+1\right)}{\left(n+1\right)^{n}\cdot\left(n+1\right)}=e\cdot\frac{e^{n}\cdot n!}{\left(n+1\right)^{n}}=e\cdot\frac{e^{n}\cdot n!}{n^{n}\left(1+\frac{1}{n}\right)^{n}}=\frac{e}{\left(1+\frac{1}{n}\right)^{n}}\cdot\frac{e^{n}\cdot n!}{n^{n}} $
now use I.H : $ \frac{e}{\left(1+\frac{1}{n}\right)^{n}}\cdot\frac{e^{n}\cdot n!}{n^{n}}\geq\frac{e}{\left(1+\frac{1}{n}\right)^{n}} $ the sequence $ \left(1+\frac{1}{n}\right)^{n} $ is increasing and converges to e. so for every $ n \in N $ we can say that $ \frac{e}{\left(1+\frac{1}{n}\right)^{n}}\geq1 $ (e is bigger) and eventually we have $ \sum_{n=1}^{\infty}\frac{e^{n}n!}{n^{n}}\geq\sum_{n=1}^{\infty}1 $ therefore the series diverge due to Convergence tests. also the limit of the sequence $ \frac{e^n \cdot n!}{n^n} $ isnt 0, then for a=-e the limit wont be 0 also, so the infinite series diverge for both cases