Zeta Function at Negative Integers

In my previous post I mentioned that there is a connection between analytic expansion value of zeta function at negative integers and a geometrical method. It turns out that the proof is very simple. We know that $$\zeta(-n)=-\frac{B_{n+1}}{n+1}$$
where $B_n$ is the $n$th Bernoulli number. Bernoulli numbers are defined as
$$B_0=1$$
$$B_m=-\frac{1}{m+1}\sum_{k=0}^{m-1}{{m+1}\choose{k}}B_k \hspace{12 mm} m \geq 1$$
From Faulhaber's formula we know that,
$$S_k(n)=\sum_{i=1}^{n-1}i^k=\frac{1}{k+1}\sum_{j=0}^{k}{{k+1}\choose{j}}B_jn^{k+1-j}$$

Integrating $S_k(x)$ from $-1$ to $0$,
$$\int_{-1}^{0}S_k(x)dx=\frac{1}{k+1}\sum_{j=0}^{k}{{k+1}\choose{j}}B_j\int_{-1}^{0}x^{k+1-j}dx$$
$$=\frac{1}{k+1}\sum_{j=0}^{k}{{k+1}\choose{j}}\frac{B_j}{k+2-j}$$
$$=\frac{1}{(k+1)(k+2)}\sum_{j=0}^{k}{{k+2}\choose{j}}B_j$$
$$=-\frac{B_{k+1}}{k+1}=\zeta(-k)$$

So the small area enclosed by $S_k(x)$ around the $x$-axis is exactly equal to $\zeta(-k)$.

Derivation of the $\zeta(-n)$ can be found here. We need to know the value of zeta function for even positive integers to derive it. A nice proof for even positive integers is described here.

Thanks for reading :)