Intereting Posts

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Open covers by simply connected sets and fundamental group
Flip two coins, if at least one is heads, what is the probability of both being heads?
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Characterizing injective polynomials

I could find the role of Striling Numbers in the natural logarithm function as follows

$$x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x})^3+…\frac{S_{n}}{n!}(\frac{x-1}{x})^n$$

Where $S_n$= absolute Striling Numbers of first kind (0,1,3,11,50,274….)

this series numerically checked without any problem, but I need the proving. Any help?

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$$\log(1-x)=-\sum_{n=1}^\infty \frac{1}{n}x^n$$

so

$$\log(x)=-\log\frac{1}{x}=-\log(1-\frac{x-1}{x})=\sum_{n=1}^\infty \frac{1}{n}(\frac{x-1}{x})^n$$

And

$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$

so

$$x=\frac{1}{1-\frac{x-1}{x}}=\sum_{n=0}^\infty (\frac{x-1}{x})^n$$

and thus

$$x\log(x)=(\sum_{n=1}^\infty \frac{1}{n}(\frac{x-1}{x})^n)\cdot \sum_{m=0}^\infty (\frac{x-1}{x})^m = \sum_{n=1}^\infty a_n(\frac{x-1}{x})^n$$

where

$$a_n=\sum_{k=0}^{n-1} \frac{1}{n-k}=1+\frac{1}{2}+\frac{1}{3}+….\frac{1}{n}=\frac{S_n}{n!}$$

So the result is

$$x\log(x)=\sum_{n=1}^\infty \frac{S_n}{n!}(\frac{x-1}{x})^n$$

we will need the fact that $$n!H_n = n!\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} \right) = S(n,2)\\ \text{ where } S(n,2) \text{ is the Stirlings number of the first kind}$$

let me make a change of variable $u= \dfrac{x-1}{x}, x = \dfrac{1}{1-u}.$ then

$\begin{align} x\ln x &= \dfrac{1}{1-u} \ln \left(\dfrac{1}{1-u}\right) = -\dfrac{1}{1-u} \ln (1-u)\\

&=\left(1+u+u^2 + \cdots\right)\left(u+\frac{u^2}{2}+\frac{u^3}{3} + \cdots\right)\\

&=u+(\frac{1}{2}+1)u^2+(\frac{1}{3}+\frac{1}{2}+1)u^3+(\frac{1}{4}+\frac{1}{3} + \frac{1}{2}+1)u^4+\cdots\\

&=u+H_2u^2+H_3u^3+H_4u^4+\cdots\\

&=u+\frac{S(2,2)}{2!}u^2+\frac{S(3,2)}{3!}u^3+\frac{S(4,2)}{4!}u^4+\cdots

+\frac{S(n,2)}{n!}u^n+\cdots\\

\end{align}$

We have

$$x\log(x) = x[-\log(1/x)] = x\left[-\log\left(1 – \frac{x-1}{x}\right)\right] = x\sum_{m = 0}^\infty \frac{1}{m+1}\left(\frac{x-1}{x}\right)^{m+1},$$

and $$x = \dfrac{1}{\frac{1}{x}} = \dfrac{1}{1 – \frac{x-1}{x}} = \sum_{m = 0}^\infty \left(\frac{x – 1}{x}\right)^m.$$ Thus

$$x\log(x) = \sum_{m = 0}^\infty \left(\frac{x-1}{x}\right)^m \sum_{m = 0}^\infty \frac{1}{m}\left(\frac{x-1}{x}\right)^m = \sum_{m = 0}^\infty \sum_{k = 0}^m \frac{1}{k+1}\left(\frac{x-1}{x}\right)^{m+1},$$

which is the same as $$\sum_{m = 0}^\infty \frac{S_m}{m!}\left(\frac{x-1}{x}\right)^{m+1}$$

Suppose we seek to show that

$$x\log x = \sum_{n=1}^\infty \frac{1}{n!} \left[n+1\atop 2\right]

\left(\frac{x-1}{x}\right)^n.$$

Recall the species equation for decompositions into disjoint cycles:

$$\mathfrak{P}(\mathcal{U}\mathfrak{C}(\mathcal{Z}))$$

which gives the generating function

$$G(z, u) = \exp\left(u\log\frac{1}{1-z}\right)$$

so that

$$\left[n+1\atop 2\right]

= \frac{1}{2} (n+1)! [z^{n+1}] \left(\log\frac{1}{1-z}\right)^2.$$

Substitute this into the sum to get

$$\frac{1}{2} \sum_{n=1}^\infty \frac{1}{n!}

(n+1)! \left(\frac{x-1}{x}\right)^n

[z^{n+1}] \left(\log\frac{1}{1-z}\right)^2

\\ = \frac{1}{2} \sum_{n=1}^\infty

(n+1) \left(\frac{x-1}{x}\right)^n

[z^{n+1}] \left(\log\frac{1}{1-z}\right)^2.$$

Now $\left(\log\frac{1}{1-z}\right)^2$ starts at $z^2$ so that we may

include the value for $n=0$ in the sum (zero contribution) to get

$$\frac{1}{2} \sum_{n=0}^\infty

(n+1) \left(\frac{x-1}{x}\right)^n

[z^{n+1}] \left(\log\frac{1}{1-z}\right)^2.$$

What we have here is an *annihilated coefficient extractor* that

evaluates to

$$\frac{1}{2}

\left. \frac{d}{dz} \left(\log\frac{1}{1-z}\right)^2

\right|_{z=(x-1)/x}.$$

This is

$$\left. \frac{1}{1-z} \log\frac{1}{1-z}

\right|_{z=(x-1)/x}$$

which finally yields

$$\frac{1}{1-(x-1)/x} \log\frac{1}{1-(x-1)/x}

= \frac{x}{x-(x-1)} \log\frac{x}{x-(x-1)}

\\ = x \log x.$$

There is another *annihilated coefficient extractor* at this

MSE link I and another one at this

MSE link II and at this MSE link III.

Admittedly the same is done in one form or another in previous answers, I just wanted to make it as concise as possible.

Substituting $x=\frac1{1-y}$, what we have to prove is$$(1+y+…+y^n+…)(y+\frac{y^2}2+…+\frac{y^n}n+…)=y+\frac3{2!}y^2+…+\frac{S_n}{n!}y^n+…,$$i. e. $n!(1+\frac12+…+\frac1n)=S_n$.

Now what you denote by $S_n$ is usually denoted $|s(n+1,2)|$ or $\left[\matrix{n+1\\2}\right]$, and the needed equality is explicitly given e. g. in Wikipedia.

Just for fun, here is a quick combinatorial proof. By definition $S_n$ is the number of permutations of $n+1$ consisting of exactly two (disjoint) cycles. To name such a permutation means to name a subset of $\{1,…,n+1\}$ of size, say, $k$, with $0<k<n+1$, together with cyclic ordering of this subset and of its complement.

Now the number of subsets of size $k$ is $\binom{n+1}k$, while the number of cyclic orderings of a set of size $\ell$ is $(\ell-1)!$, so we get $\binom{n+1}k(k-1)!(n-k)!=\frac{(n+1)!}{k(n+1-k)}=n!\left(\frac1k+\frac1{n+1-k}\right)$. We then have to sum over $1\le k\le n$ and divide by 2 (since we have counted each disjoint pair of cycles twice, first at $k$ and then at $n+1-k$). The result is $n!H_n$.

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