# Evaluating $\lim_{n\to \infty}\frac{n!}{n^n}\left(\sum_{k=0}^{n}\frac{n^k}{k!}-\sum_{k=n+1}^{\infty}\frac{n^k}{k!}\right)$ [closed]

How do you evaluate the following limit: $$\lim_{n\to \infty}\frac{n!}{n^n}\left(\sum_{k=0}^{n}\frac{n^k}{k!}-\sum_{k=n+1}^{\infty}\frac{n^k}{k!}\right)$$

• What have you done to try the problem so far Oct 9, 2019 at 23:43
• I tried to evaluate the first term n!/n^n and got 0, but I don't know how to evaluate the difference between the two series. Oct 9, 2019 at 23:47
• Notice that the term with the summations is very closely related to the expansion of e^n. Oct 10, 2019 at 3:12

Here is a partial solution towards the following claim:

Claim. We have $$A_n := \frac{n!}{n^n} \left( \sum_{k=0}^{n} \frac{n^k}{k!} - \sum_{k=n+1}^{\infty} \frac{n^k}{k!} \right) \xrightarrow[n\to\infty]{} \frac{4}{3}.$$

This claim is supported by numerical experiments, and we aim at justifying this claim.

Step 1. Notice that

$$\sum_{k=0}^{n} \frac{n^k}{k!} = e^n \int_{n}^{\infty} \frac{x^n e^{-x}}{n!} \, \mathrm{d}x \stackrel{(x=n+\sqrt{n} u)}=\frac{n^{n+\frac{1}{2}}}{n!} \int_{0}^{\infty} \left(1 + \frac{u}{\sqrt{n}}\right)^{n}e^{-\sqrt{n}u} \, \mathrm{d}u$$

and similarly

$$\sum_{k=n+1}^{\infty} \frac{n^k}{k!} = e^n \int_{0}^{n} \frac{x^{n}e^{-x}}{n!} \, \mathrm{d}x \stackrel{(x=n-\sqrt{n} u)}=\frac{n^{n+\frac{1}{2}}}{n!} \int_{0}^{\sqrt{n}} \left(1 - \frac{u}{\sqrt{n}}\right)^{n}e^{\sqrt{n}u} \, \mathrm{d}u.$$

Combining altogether, we get

\begin{align*} A_n &:= \frac{n!}{n^n} \left( \sum_{k=0}^{n} \frac{n^k}{k!} - \sum_{k=n+1}^{\infty} \frac{n^k}{k!} \right) \\ &= \sqrt{n} \left( \int_{0}^{\infty} \left(1 + \frac{u}{\sqrt{n}}\right)^{n}e^{-\sqrt{n}u} \, \mathrm{d}u - \int_{0}^{\sqrt{n}} \left(1 - \frac{u}{\sqrt{n}}\right)^{n}e^{\sqrt{n}u} \, \mathrm{d}u \right). \tag{1} \end{align*}

This expression will be our starting point for examining the limiting behavior of $$A_n$$.

Step 2. To make the analysis simpler, set $$f_n, g_n : [0,\infty) \to \mathbb{R}$$ by

$$f_n(u) = \left(1 + \frac{u}{\sqrt{n}}\right)^{n}e^{-\sqrt{n}u}, \qquad g_n(x) = \left(1 - \frac{u}{\sqrt{n}}\right)^{n}e^{\sqrt{n}u}\mathbf{1}_{[0,n]}(u).$$

We will make heavy use of the following observation on $$f_n$$ and $$g_n$$:

• $$f_n(u) \downarrow e^{-u^2/2}$$ and $$g_n(u) \uparrow e^{-u^2/2}$$ as $$n\to\infty$$ for each fixed $$u \geq 0$$.

Indeed, this can be easily proved by allowing $$n$$ to take any positive real value and investigating first two derivatives of $$\log f_n$$ and $$\log g_n$$. Now using this,

$$\sqrt{n} \int_{\sqrt{n}/2}^{\infty} f_n(u) \, \mathrm{d}u \leq \sqrt{n} \int_{\sqrt{n}/2}^{\infty} f_{1}(u) \, \mathrm{d}u = \frac{1}{2}\sqrt{n}(\sqrt{n}+4)e^{-\sqrt{n}/2}$$

and

$$\sqrt{n} \int_{\sqrt{n}/2}^{\sqrt{n}} g_n(u) \, \mathrm{d}u \leq \sqrt{n} \int_{\sqrt{n}/2}^{\infty} e^{-u^2/2} \, \mathrm{d}u \leq 2 \int_{\sqrt{n}/2}^{\infty} u e^{-u^2/2} \, \mathrm{d}u \leq 2 e^{-n/8}.$$

So we may truncate both integral at $$\sqrt{n}/2$$ to write

$$A_n = \sqrt{n} \int_{0}^{\sqrt{n}/2} (f_n(u) - g_n(u)) \, \mathrm{d}u + o(1).$$

Finally, let $$0 \leq u \leq \sqrt{n}/2$$ and notice that

\begin{align*} \frac{g_n(u)}{f_n(u)} &= \exp\left( 2\sqrt{n}u - 2n \sum_{k=1,3,5,\cdots} \frac{u^k}{k n^{k/2}} \right) \\ &\geq \exp\left( - 2n \sum_{k=3,5,7,\cdots} \frac{u^3}{k n^{3/2}} \left(\frac{1}{2}\right)^{k-3} \right) \\ &= \exp\left( - \frac{c u^3}{\sqrt{n}} \right) \end{align*}

for some absolute constant $$c \in (0, \infty)$$, and so,

$$\sqrt{n}\left(f_n(u) - g_n(u) \right) \leq \sqrt{n} \left( 1 - e^{-c u^3/\sqrt{n}} \right) f_n(u) \leq c u^3 f_n(u)$$

uniformly in $$0 \leq u \leq \sqrt{n}/2$$. (In the last step, we utilized the inequality $$1 - e^{-x} \leq x$$.) Therefore by the dominated convergence theorem,

\begin{align*} \lim_{n\to\infty} A_n &= \int_{0}^{\infty} \lim_{n\to\infty} \sqrt{n}\left(f_n(u) - g_n(u) \right) \mathbf{1}_{[0,\sqrt{n}/2]}(u) \, \mathrm{d}u \\ &= \int_{0}^{\infty} \frac{2}{3}u^3 e^{-u^2/2} \, \mathrm{d}u \\ &= \frac{4}{3} \end{align*}

as desired. $$\square$$

• I do not know where I made a mistake but I found a limit equal to $\frac 43$ also supported by numerical evaluations. Oct 10, 2019 at 5:26
• @ClaudeLeibovici, Indeed I made a mistake when rephrasing the sum by the integral. Now it is fixed, and the correct integral representation allowed me to produce the correct answer as you conjectured. Thank you! Oct 10, 2019 at 15:54

Write the inner sums as

$$2\sum_{k=0}^{n}\frac{n^k}{k!}-e^n$$ and look up Ramanujan.

(Done on my phone so I can't do much more right now.)