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

  • $\begingroup$ What have you done to try the problem so far $\endgroup$
    – QC_QAOA
    Oct 9, 2019 at 23:43
  • $\begingroup$ 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. $\endgroup$
    – John He
    Oct 9, 2019 at 23:47
  • $\begingroup$ Notice that the term with the summations is very closely related to the expansion of e^n. $\endgroup$
    – 63677
    Oct 10, 2019 at 3:12

2 Answers 2


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,

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


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

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. $\square$

  • $\begingroup$ I do not know where I made a mistake but I found a limit equal to $\frac 43$ also supported by numerical evaluations. $\endgroup$ Oct 10, 2019 at 5:26
  • $\begingroup$ @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! $\endgroup$ 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.)


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