Proved that question:

$S\left(n\right)=\sum _{k\ge 1}\frac{n!}{k(n-k)!{n}^{k}}=\sum _{k\ge 1}\frac{{n}^{\underset{\u2015}{k}}}{k{n}^{k}}\approx \frac{1}{2}\mathrm{log}\left(n\right)$

Bronson Olson
2022-05-02
Answered

Proved that question:

$S\left(n\right)=\sum _{k\ge 1}\frac{n!}{k(n-k)!{n}^{k}}=\sum _{k\ge 1}\frac{{n}^{\underset{\u2015}{k}}}{k{n}^{k}}\approx \frac{1}{2}\mathrm{log}\left(n\right)$

You can still ask an expert for help

Cloplerotly1gu

Answered 2022-05-03
Author has **11** answers

Here is a rough argument. Filling in the details should be elementary, but challenging. I find myself reusing several ideas from this earlier question.

The sum is

$\sum _{k=1}^{n}\frac{1}{k}(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{k}{n})$

As discussed in the earlier answer,

$(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{k}{n})\approx {e}^{-\frac{{k}^{2}}{2n}}$

So the sum is roughly

$\sum _{k\ge 1}\frac{1}{k}{e}^{-\frac{{k}^{2}}{2n}}=\sum _{k\ge 1}\frac{1}{\sqrt{n}}\frac{\sqrt{n}}{k}{e}^{-\frac{{\left(\frac{k}{\sqrt{n}}\right)}^{2}}{2}}$

This is a Riemann sum approximation (with interval size$\frac{1}{\sqrt{n}}$ ) to the integral

Let$\left[x<1\right]$ be 1 for $x<1$ and 0 $x\ge 1$ . Then the integral is

${\int}_{\frac{1}{\sqrt{n}}}^{1}\frac{dx}{x}+{\int}_{\frac{1}{\sqrt{n}}}^{\mathrm{\infty}}\frac{{e}^{-\frac{{x}^{2}}{2}}-\left[x<1\right]}{x}dx$

The first integral is$\left(\frac{1}{2}\right)\mathrm{log}n$ . The second approaches ${\int}_{0}^{\mathrm{\infty}}\frac{{e}^{-\frac{{x}^{2}}{2}}-\left[x<1\right]}{x}dx$ , which is convergent.

So our final integral is$\left(\frac{1}{2}\right)\mathrm{log}n+O\left(1\right)$

The sum is

As discussed in the earlier answer,

So the sum is roughly

This is a Riemann sum approximation (with interval size

Let

The first integral is

So our final integral is

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