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In this post here, I found a proof that I was intrigued by. However, I have several questions about it. Here is the proof:

A proof I found a while ago entirely relies on creative telescoping. Since $\frac{1}{n^2}-\frac{1}{n(n+1)}=\frac{1}{n^2(n+1)}$,

$$\begin{eqnarray*} \sum_{n\geq m}\frac{1}{n^2}&=&\sum_{n\geq m}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\sum_{n\geq m}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\\&+&\frac{1}{6}\sum_{n\geq m}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)-\frac{1}{6}\sum_{n\geq m}\frac{1}{n^3(n+1)^3}\tag{1}\end{eqnarray*} $$ hence: $$ \psi'(m)=\sum_{n\geq m}\frac{1}{n^2}\leq \frac{1}{m}+\frac{1}{2m^2}+\frac{1}{6m^3}\tag{2}$$ and in a similar fashion: $$ \psi'(m) \geq \frac{1}{m}+\frac{1}{2m^2}+\frac{1}{6m^3}-\frac{1}{30m^5}.\tag{3}$$ Integrating twice, we have that $\log\Gamma(m)$ behaves like: $$ \log\Gamma(m)\approx\left(m-\frac{1}{2}\right)\log(m)-\color{red}{\alpha} > m+\color{blue}{\beta}+\frac{1}{12m}\tag{4}$$ where $\color{red}{\alpha=1}$ follows from $\log\Gamma(m+1)-\log\Gamma(m)=\log m$.

That gives Stirling's inequality up to a multiplicative constant.

$\color{blue}{\beta=\log\sqrt{2\pi}}$ then follows from Legendre's duplication formula and the well-known identity:

$$ \Gamma\left(\frac{1}{2}\right)=2 \int_{0}^{+\infty}e^{-x^2}\,dx = \sqrt{\pi}.\tag{5}$$

  1. How did we get from $\frac{1}{n^2}-\frac{1}{n(n+1)}=\frac{1}{n^2(n+1)}$ to all the summation stuff?

  2. What is $\psi$? Where did it come from?

  3. Where did the $\frac{1}{30m^5}$ term come from?

  4. How does integrating twice get us to that?

  5. How do we get from $\log\Gamma(x)$ to $n! \approx \sqrt{2\pi n}\, n^n e^{-n}$?

Basically, explain it in such a way that a brussel sprout would understand it.

Thanks for any and all help!

D.R.
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1 Answers1

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1. $\sum_{n\geq 1}\frac{1}{n(n+1)(n+2)}$ and stuffs like that are simple to compute since they are telescopic series. $\sum_{n\geq 1}\frac{1}{n^2}$ does not have an obvious telescopic structure, but we may approximate $\frac{1}{n^2}$ with a telescopic term, $\frac{1}{n(n+1)}$, then approximate the difference $\frac{1}{n^2}-\frac{1}{n(n+1)}$ with a telescopic term and so on, till meeting the required accuracy;

2. $\psi$ is the digamma function, i.e. $\psi(x)=\frac{d}{dx}\log\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)}$. By the Weierstrass product for the $\Gamma$ function, $\psi'$ is related with $\sum_{n\geq N}\frac{1}{n^2}$. So, if we are able to approximate $\psi'$ with a decent accuracy, we are also able to approximate $\psi$ by termise integration, then $\log\Gamma$ by termwise integration again, then $\Gamma$ by exponentiation. That is the main idea of the approach. By integrating twice we lose track of two constants, but that is not too bad since we may easily recover them at the end from the functional identities fulfilled by the $\Gamma$ function;

3. The quintic term comes from approximating $\frac{1}{n^3(n+1)^3}$ with a telescopic term of the form $\frac{C}{n^5}-\frac{C}{(n+1)^5}$;

4. Already explained in 2.;

5. Already explained in 2.: $n!=\Gamma(n+1)=\exp\log\Gamma(n+1)$.

Jack D'Aurizio
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