The Zeta Function:Finding a general formula Part 3

The Cauchy Product Formula tells us that

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

,where cj=k=0jak.bjkc_j=\displaystyle\sum_{k=0}^{j} a_k.b_{j-k}

From last time we got

1=n=0βnn!.zn.n=0zn(n+1)!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=j=0k=0jβkk!.(jk+1)!zj1=\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)!(j+1)! and divide by that in the inner sum.But when doing that, we get a binomial in the right sum

So

1=j=01(j+1)!k=0j(j+1k)βkzk1=\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 β0\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

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

Thus we can find a relation between the βk\beta_k's and , as we know β0=1\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 zez1\displaystyle\frac{z}{e^z-1}.

So

π.s.cotan(π.s)=π.i.s+2π.i.se2π.i.s1=π.i.s+n=0βnn!(2π.i.s)n\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 xcotanxx cotan x is even, thus all the powers in its series must be even, so β2k+1=0\beta_{2k+1}=0 for k>0k>0.

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

π.s.cotan(π.s)=π.i.s+β00!+β11!.2π.i.s+n=2βnn!(2π.i.s)n\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 β1=0.5\beta_1=-0.5, so the first and the third cancel.Also, β2k+1=0\beta_{2k+1}=0, so we can again rewrite our sum

π.s.cotan(π.s)=1+n=1β2n2n!(2π.i.s)2n\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 12-\frac{1}{2} in the sum so we get a -2 up front.

We get

π.s.cotan(π.s)=12n=112.β2n2n!(2π.i.s)2n\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

π.s.cotan(π.s)=12.n=1ζ(2n).s2n\displaystyle \pi.s.cotan(\pi.s)=1-2.\sum_{n=1}^{\infty} \zeta (2n).s^{2n}

So comparing the coefficients we get that

12.β2n2n!(2π.i.s)2n=ζ(2n).s2n-\frac{1}{2}.\frac{\beta_{2n}}{2n!}(2\pi.i.s)^{2n} = \zeta(2n).s^{2n}

So, finally

ζ(2n)=(1)n+1β2n2n!.22n1.π2n\displaystyle\boxed{\zeta(2n)=(-1)^{n+1}\frac{\beta_{2n}}{2n!}.2^{2n-1}.\pi^{2n}}

Note by Bogdan Simeonov
5 years, 4 months ago

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