The Cauchy Product Formula tells us that

\(\displaystyle(\sum_{n=0}^{\infty} a_n )(\sum_{n=0}^{\infty} b_n )=\sum_{j=0}^{\infty} c_j\)

,where \(c_j=\displaystyle\sum_{k=0}^{j} a_k.b_{j-k}\)

From last time we got

\(1=\displaystyle\sum_{n=0}^{\infty}\frac{\beta_n}{n!}.z^n.\sum_{n=0}^{\infty} \frac{z^{n}}{(n+1)!}\)

So using the formula from above we get

\(1=\displaystyle\sum_{j=0}^{\infty}\sum_{k=0}^{j} \frac{\beta_k}{k!.(j-k+1)!}z^j\)

Since j does not depend on k, we can multiply the rightmost sum by \((j+1)!\) and divide by that in the inner sum.But when doing that, we get a binomial in the right sum

So

\(1=\displaystyle\sum_{j=0}^{\infty}\frac{1}{(j+1)!}\sum_{k=0}^{j} \binom{j+1}{k}\beta_kz^k\)

When j=0, the output is 1(because \(\beta_0\) must be 1), so the sum of all the others is exactly 0.But we defined z not to be equal to 0, so

\(\displaystyle\sum_{k=0}^{j} \binom{j+1}{k}\beta_k=0\).

Thus we can find a relation between the \(\beta_k\)'s and , as we know \(\beta_0=1\), we can find all terms in the sequence (these numbers are called Bernoulli numbers).

So now we have found the terms in the infinite series for \(\displaystyle\frac{z}{e^z-1}\).

So

\(\displaystyle\pi.s.cotan(\pi.s)=\pi.i.s + \frac{2\pi.i.s}{e^{2\pi.i.s}-1}=\pi.i.s + \sum_{n=0}^{\infty}\frac{\beta_n}{n!}(2\pi.i.s)^n\).

Note that the function \(x cotan x\) is even, thus all the powers in its series must be even, so \(\beta_{2k+1}=0\) for \(k>0\).

Now let's rewrite that sum of ours, starting from n=2.

\(\displaystyle\pi.s.cotan(\pi.s)=\pi.i.s + \frac{\beta_0}{0!}+\frac{\beta_1}{1!}.2\pi.i.s+\sum_{n=2}^{\infty}\frac{\beta_n}{n!}(2\pi.i.s)^n\)

But \(\beta_1=-0.5\), so the first and the third cancel.Also, \(\beta_{2k+1}=0\), so we can again rewrite our sum

\(\displaystyle \pi.s.cotan(\pi.s)= 1+\sum_{n=1}^{\infty}\frac{\beta_{2n}}{2n!}(2\pi.i.s)^{2n}\)

Now let's put a \(-\frac{1}{2}\) in the sum so we get a -2 up front.

We get

\(\displaystyle \pi.s.cotan(\pi.s)= 1-2\sum_{n=1}^{\infty} -\frac{1}{2}.\frac{\beta_{2n}}{2n!}(2\pi.i.s)^{2n} \)

From Part 2 we know that

\(\displaystyle \pi.s.cotan(\pi.s)=1-2.\sum_{n=1}^{\infty} \zeta (2n).s^{2n}\)

So comparing the coefficients we get that

\(-\frac{1}{2}.\frac{\beta_{2n}}{2n!}(2\pi.i.s)^{2n} = \zeta(2n).s^{2n}\)

So, finally

\(\displaystyle\boxed{\zeta(2n)=(-1)^{n+1}\frac{\beta_{2n}}{2n!}.2^{2n-1}.\pi^{2n}}\)

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