Ytukyg

I found this formula. Is it correct?

For $ninBbb N, ngeq2$,
$$zeta(2n)=frac{2npi^{2n}}{Gamma(2n+2)}+sum_{k=0}^{n-2}(-1)^{k-n}frac{pi^{2n-2k-2}}{Gamma(2n-2k)}zeta(2k+2)$$

Here's my proof.

Assume $mgeq2$ is an even number. We start by finding the Fourier series expansion for $x^m$. We define
$$a_k(m)=frac1piint_{-pi}^{pi}x^mcos(kx)mathrm dx$$
$$b_k(m)=frac1piint_{-pi}^{pi}x^msin(kx)mathrm dx$$
Hence we know that
$$x^m=frac{a_0(m)}2+sum_{kgeq1}a_k(m)cos(kx)+b_k(m)sin(kx)$$
And because $m$ is even,
$$x^m=frac{pi^m}{m+1}+sum_{kgeq1}a_k(m)cos(kx)+b_k(m)sin(kx)$$
Next, we start on $b_k(m)$. We make the substitution $x=-t$, and after a little algebra,
$$b_k(m)=0$$
This is because $m$ is even, and thus $(-x)^m=x^m$. Hence we update our series
$$x^m=frac{pi^m}{m+1}+sum_{kgeq1}a_k(m)cos(kx)$$
For $a_k(m)$, I integrated by parts twice to find
$$a_k(m)=(-1)^{k}frac{2mpi^{m-2}}{k^2}-frac{m(m-1)}{k^2}a_k(m-2)$$
Which brought me to the formula
$$a_k(m)=sum_{v=0}^{frac{m}2-1}(-1)^{k+v}frac{2pi^{m-2-2v}}{k^{2+2v}}prod_{i=1}^{2v+1}(m+1-i)$$
Then, plugging $x=pi$ into the Fourier series,
$$pi^m=frac{pi^m}{m+1}+sum_{kgeq1}(-1)^ksum_{v=0}^{frac{m}2-1}(-1)^{k+v}frac{2pi^{m-2-2v}}{k^{2+2v}}prod_{i=1}^{2v+1}(m+1-i)$$
$$frac{mpi^m}{m+1}=sum_{kgeq1}sum_{v=0}^{frac{m}2-1}(-1)^{v}frac{2pi^{m-2-2v}}{k^{2+2v}}prod_{i=1}^{2v+1}(m+1-i)$$
$$frac{mpi^m}{m+1}=sum_{v=0}^{frac{m}2-1}(-1)^{v}2pi^{m-2-2v}prod_{i=1}^{2v+1}(m+1-i)sum_{kgeq1}frac1{k^{2+2v}}$$
$$frac{mpi^m}{m+1}=sum_{v=0}^{frac{m}2-1}(-1)^{v}2pi^{m-2-2v}zeta(2v+2)prod_{i=1}^{2v+1}(m+1-i)$$
Defining
$$c_v(m)=(-1)^v2prod_{i=1}^{2v+1}(m+1-i)$$
We have
$$frac{mpi^m}{m+1}=sum_{v=0}^{frac{m}2-1}pi^{m-2-2v}c_v(m)zeta(2v+2)$$
$$frac{mpi^m}{m+1}=c_{m/2-1}(m)zeta(m)+sum_{v=0}^{frac{m}2-2}pi^{m-2-2v}c_v(m)zeta(2v+2)$$
$$c_{m/2-1}(m)zeta(m)=frac{mpi^{m}}{m+1}-sum_{v=0}^{frac{m}2-2}pi^{m-2-2v}c_v(m)zeta(2v+2)$$
I then replace $m$ with $2n$ and $v$ with $k$ (because I like $k$ better):
$$c_{n-1}(2n)zeta(2n)=frac{2npi^{2n}}{2n+1}-sum_{k=0}^{n-2}pi^{2n-2-2k}c_k(2n)zeta(2k+2)$$
$$zeta(2n)=frac{2npi^{2n}}{(2n+1)c_{n-1}(2n)}-sum_{k=0}^{n-2}pi^{2n-2-2k}frac{c_k(2n)}{c_{n-1}(2n)}zeta(2k+2)$$
At this point I plugged some things into Wolfram Alpha and got
$$zeta(2n)=frac{2npi^{2n}}{Gamma(2n+2)}+sum_{k=0}^{n-2}(-1)^{k-n}frac{pi^{2n-2k-2}}{Gamma(2n-2k)}zeta(2k+2)$$

I have not seen this formula anywhere, and I don't have Mathematica or Wolfram Pro, so the only way for me to feel confident in this formula is by having you look at it. I don't think I broke any rules, so is this formula valid? Are there any alternate forms for it? Thanks.

It is correct, but it can be proved with considerably fewer efforts. Since Euler's work it is well-known that
$$ sum_{ngeq 1} zeta(2n) z^{2n} = frac{1-pi z cot(pi z)}{2} $$
hence the values of the $zeta$ function at the positive even integers are related to the coefficients of the Maclaurin series of $pi z cot(pi z)$, or $zcoth z$, or $frac{z}{e^z-1}$. So we have an explicit relation between $zeta(2n)$ and the Bernoulli number $B_{2n}$. Since the reciprocal of the (exponential) generating function of Bernoulli numbers is
$$ frac{e^z-1}{z} = sum_{ngeq 0}frac{z^{n}}{(n+1)!} $$
by considering the Cauchy product between $frac{e^z-1}{z}$ and $frac{z}{e^z-1}$ we may easily derive the recurrence relation for Bernoulli numbers (with even index). This recurrence, translated in terms of $zeta(2n)$, is exactly your identity.