# The convergence of $\sum_{n=1}^\infty (-1)^n\left(\frac{n}{e}\right)^n\frac{1}{n!}$

I'm trying to figure out if the $$\sum_{n=1}^\infty (-1)^n\left(\frac{n}{e}\right)^n\frac{1}{n!}$$ converges or not. I've tried the Leibnitz test for alternating series, but it leads to Stirling's formula and I was wondering if there's any other way so I could avoid using it. I'll be grateful for any idea.

• Check to see if the inside of the sum converges to $0$. If it does not, then the sum is divergent. Nov 12, 2019 at 15:58
• @Anonymous But I think it does go to $0$. Nov 12, 2019 at 16:02
• @Milten My bad, you're right. Nov 12, 2019 at 16:10
• Nov 12, 2019 at 16:12

$$a_n = \frac{n^n e^{-n}}{n!}$$ is a positive and decreasing sequence with limit zero, hence the series is convergent by Leibniz rule. $$\text{decreasing}:\qquad \frac{a_{n+1}}{a_n} = \frac{1}{e}\left(1+\frac{1}{n}\right)^n<1.$$ $$\text{convergent to zero}:\left\{ \begin{eqnarray*}\log(n!)&=&\sum_{k=1}^{n}\log(k)=n\log n-\sum_{k=1}^{n-1}k\log\left(1+\frac{1}{k}\right)\\&\geq &n\log n-\sum_{k=1}^{n-1}k\left(\frac{1}{k}-\frac{1}{4k^2}\right)\\&\geq &n\log n-n+\frac{1}{4}\log n.\end{eqnarray*}\right.$$ By the Lagrange inversion theorem (see 1 and 2) we have $$-\frac{W(x)}{1+W(x)} = \sum_{n\geq 1}\frac{(-1)^{n}n^{n}}{n!}\,x^n$$ for any $$x$$ sufficiently close to the origin, with $$W(x)$$ being Lambert's function, i.e. the inverse function of $$x e^x$$.
It follows that $$\sum_{n\geq 1}\frac{(-1)^n n^n}{e^n n!} = -\frac{W(1/e)}{1+W(1/e)}$$ and by Newton's method the value of the series is approximately $$-0.2178117$$.

• @BarryCipra: that is easily derived from Stirling's approximation $n!\approx n^n e^{-n}\sqrt{2\pi n}$. Nov 12, 2019 at 16:34
• And is there any other way how to show that the limit of sequence is 0 if I couldn't use the Stirling's approximation? Thanks for the quick answer @Jack D'Aurizio Nov 12, 2019 at 16:48
• @BarryCipra: I have added a proof of $a_n\to 0$ independent from Stirling's approximation. It is enough to exploit summation by parts and the inequality $\log(1+x)\leq x-\frac{x^2}{4}$ on $(0,1]$. Nov 12, 2019 at 18:09

Too long for a comment but not a (complete) answer:

Notice that there is a theorem due to Stirling asserting that for big $$n$$ one has:

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

So, in particular, for big $$n$$, the term of your sum is $$(-1)^n\frac{1}{\sqrt{2n\pi}}$$ which tells us that it will for sure converge (since it is an alternating sum of decreasing and tending to zero values).

It is possible to show convergence while avoiding Leibniz', and other, convergence tests. However, I've used a slightly convoluted route. I'm assuming that's what you meant, not avoiding Stirling's approximation. First, combining odd and even terms with $$b_n=a_{2n}-a_{2n-1}$$, the series equals

\begin{aligned}S&=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}\left(\frac{n}{e}\right)^{n}}{n!} \\ &=\sum_{n=1}^{\infty}\frac{\left(2n\right)^{2n-1}-e\cdot\left(2n-1\right)^{2n-1}}{e^{2n}\cdot\left(2n-1\right)!} \end{aligned}

The factorial can be bounded with the lower bound of Stirling's approximation, $$\sqrt{2\pi}\ n^{n+\frac12}e^{-n} \le n!$$

\begin{aligned}|S|\leq \left|\frac{1}{e\sqrt{2\pi}}\sum_{n=1}^{\infty}\frac{\left(\frac{2n}{2n-1}\right)^{2n-1}-e}{\left(2n-1\right)^{\frac{1}{2}}}\right| \end{aligned}

Note the series on the right is negative so its sign is flipped by the modulus from this point. With $$\ln n \leq n-1$$, we have $$\left(\frac{2n}{2n-1}\right)^{2n-1} = e^{-\left(2n-1\right)\ln\left(1-\frac{1}{2n}\right)}\geq e^{1-\frac{1}{2n}}$$

\begin{aligned}|S|&\leq \frac{1}{\sqrt{2\pi}}\sum_{n=1}^{\infty}\frac{e^{\frac{1}{2n}}-1}{e^{\frac{1}{2n}}\left(2n-1\right)^{\frac{1}{2}}} \\ &\leq \frac{1}{\sqrt{2\pi}}\sum_{n=1}^{\infty}\frac{e^{\frac{1}{2n}}-1}{\left(2n-1\right)^{\frac{1}{2}}} \end{aligned}

As $$e^x=\frac{1}{e^{-x}}\leq\frac{1}{1-x}$$,

\begin{aligned}|S|&\leq \frac{1}{\sqrt{2\pi}}\sum_{n=1}^{\infty}\frac{1}{\left(2n-1\right)^{\frac{3}{2}}} \\ &\leq \frac{1}{\sqrt{2\pi}}\left(\sum_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}-\sum_{n=1}^{\infty}\frac{1}{\left(2n\right)^{\frac{3}{2}}}\right) \\ &\leq \frac{1-\frac{1}{2\sqrt{2}}}{\sqrt{2\pi}}\sum_{n=1}^{\infty}{n^{-3/2}} \end{aligned}

Finally, as the series on the right has a strictly decreasing summand, we have $$\sum_{n=1}^{\infty}{n^{-3/2}}\leq 1+\int_2^\infty (t-1)^{-3/2}\ \mathrm{d}t=3$$, so $$|S|\leq\frac{3}{4\sqrt{\pi}}\left(2\sqrt{2}-1\right)=0.774$$ and $$S$$ is convergent.

• I'm sorry for misunderstanding, but this is exactly what I didn't mean- I was actually trying to avoid Stirling, not Leibnitz rule :) but thank you for your time anyway Nov 13, 2019 at 15:01