What's the links between factorial,Basel Problem and Euler constant .

Question

I'm studying the factorial function and I have found the following identity :

$$\frac{\partial^2 x!}{\partial x^2}=-2\gamma+\gamma^2+\frac{\pi^2}{6}$$ At $x=1$

But the Basel problem says : $$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}.$$

And $\gamma$ constant is related to Harmonic series .

My question is how to explain these links ?

Thanks a lot for your time .

Can you add the link where you've found the identity? Since $\Gamma(n)=(n-1)!$ it follows from this duplicate, applied twice (I hope). — Dietrich Burde, Aug 29 '19 at 14:30

score 1 · Accepted Answer · answered Sep 06 '19 at 22:24

As noted in the comments, this follows from Derivative of a factorial. Assuming we extend the factorial using the Gamma function we have:

$$x!=x\cdot(x-1)!$$

$$\psi_{-1}(x+1)=\ln(x!)=\ln(x)+\psi_{-1}(x)$$

Differentiating both sides twice gives us the trigamma function:

$$\psi_1(x+1)=\psi_{-1}''(x+1)=-\frac1{x^2}+\psi_1(x)$$

but the second derivative of $\ln(g(x))$ is given by

$$\frac{\mathrm d^2}{\mathrm dx^2}\ln(g(x))=\frac{g''(x)g(x)-[g'(x)]^2}{[g(x)]^2}$$

However, using the recurrence relation gives us:

$$\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\psi_1(n)-\psi_1(n+1)=\psi_1(1)-\underbrace{\lim_{n\to\infty}\psi_1(n)}_{{}=0}=\psi_1(2)+1$$

which gives

$$\frac{\pi^2}6=(x!)''-[(x!)']^2\bigg|_{x=1}+1$$

1 Answers1