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# The Zeta Function:Finding a general formula Part 3

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

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

Note by Bogdan Simeonov
3 years, 6 months ago