Proved that question: \(\displaystyle{S}{\left({n}\right)}=\sum_{{{k}\geq{1}}}{\frac{{{n}!}}{{{k}{\left({n}-{k}\right)}!{n}^{{k}}}}}=\sum_{{{k}\geq{1}}}{\frac{{{n}^{{\underline{{{k}}}}}}}{{{k}{n}^{{k}}}}}\approx{\frac{{{1}}}{{{2}}}}{\log{{\left({n}\right)}}}\)

calcolare45pj

calcolare45pj

Answered question

2022-03-30

Proved that question:
S(n)=k1n!k(nk)!nk=k1nkknk12log(n)

Answer & Explanation

Karsyn Wu

Karsyn Wu

Beginner2022-03-31Added 17 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
k=1n1k(11n)(12n)(1kn)
As discussed in the earlier answer,
(11n)(12n)(1kn)ek22n
So the sum is roughly
k11kek22n=k11nnke(kn)22
This is a Riemann sum approximation (with interval size 1n) to the integral
Let [x<1] be 1 for x<1 and 0 x1. Then the integral is
1n1dxx+1nex22[x<1]xdx
The first integral is (12)logn. The second approaches 0ex22[x<1]xdx, which is convergent.
So our final integral is (12)logn+O(1)

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