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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.

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3 Answers 3

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$$a_n = \frac{n^n e^{-n}}{n!} $$ 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 $-0.2178117$.

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  • $\begingroup$ @BarryCipra: that is easily derived from Stirling's approximation $n!\approx n^n e^{-n}\sqrt{2\pi n}$. $\endgroup$ Nov 12, 2019 at 16:34
  • $\begingroup$ 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 $\endgroup$
    – thepotato
    Nov 12, 2019 at 16:48
  • $\begingroup$ @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 $(0,1]$. $\endgroup$ Nov 12, 2019 at 18:09
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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).

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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.

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  • $\begingroup$ 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 $\endgroup$
    – thepotato
    Nov 13, 2019 at 15:01

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