# series

• Jan 30th 2011, 10:56 AM
Random Variable
series
$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \ \zeta(2n)$
• Jan 30th 2011, 12:22 PM
chisigma
If we define the 'factorial function' as...

$\displaystyle z!= \int_{0}^{\infty} t^{z}\ e^{-t}\ dt$ (1)

... with a little of patience we 'discover' that is...

$\displaystyle \ln \frac{1}{z!}= \gamma\ z - \sum_{n=2}^{\infty} (-1)^{n} \zeta(n)\ \frac{z^{n}}{n}$ (2)

... where $\gamma$ is the 'Euler's constant'. To be continued in a succesive post...

Kind regards

$\chi$ $\sigma$
• Jan 30th 2011, 01:32 PM
Random Variable
• Jan 30th 2011, 02:06 PM
chisigma
Quote:

Originally Posted by chisigma
If we define the 'factorial function' as...

$\displaystyle z!= \int_{0}^{\infty} t^{z}\ e^{-t}\ dt$ (1)

... with a little of patience we 'discover' that is...

$\displaystyle \ln \frac{1}{z!}= \gamma\ z - \sum_{n=2}^{\infty} (-1)^{n} \zeta(n)\ \frac{z^{n}}{n}$ (2)

... where $\gamma$ is the 'Euler's constant'...

... and that is...

$\displaystyle \varphi(z)= \frac{1}{z!}\ \frac{1}{(-z)!}= \frac{\sin \pi z}{\pi z} = sinc (z)$ (3)

Combining (2) and (3) we obtain...

$\displaystyle \ln sinc(z) = -\sum_{n=1}^{\infty} \frac{\zeta(2n)}{n}\ z^{2n}$ (4)

... so that we have...

$\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \frac{\zeta(2n)}{n} = \ln sinc(i) = \ln \frac{\sinh \pi}{\pi}$ (5)

Kind regards

$\chi$ $\sigma$
• Jan 30th 2011, 03:20 PM
Random Variable
The solution I had in mine is totally different.

$\displaystyle \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} = \zeta (a+1) \Gamma(a+1), \ a>0$

so $\displaystyle \sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} \zeta (2n) = \sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} \frac{1}{\Gamma (2n)} \int^{\infty}_{0} \frac{x^{2n-1}}{e^{x}-1} \ dx$

Using the fact that $\displaystyle 1 - \cos x = \sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{(2n-1)!} \frac{x^{2n}}{2n}$ (Integrate both sides of the Maclaurin series for $\sin x$ and don't forget about the constant ).

$\displaystyle \sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} \zeta (2n) = 2 \int^{\infty}_{0} \frac{ 1- \cos x}{x(e^{x}-1)} \ dx$

$\displaystyle = 2 \int^{\infty}_{0} \frac{1-\cos x}{x} \sum_{n=1}^{\infty} e^{-nx} \ dx$

$\displaystyle = 2 \sum^{\infty}_{n=1} \int^{\infty}_{0} \int^{\infty}_{0} (1-\cos x) \ e^{-(t+n)x} \ dt \ dx$

The integrand is always positive, so you don't have to worry about something potentially going haywire when you switch the order of integration.

$\displaystyle = \2 \sum^{\infty}_{n=1} \int^{\infty}_{0} \big( \mathcal{L} \{1\} (t+n) - \mathcal{L} \{\cos x\} (t+n) \big) \ dt$

$\displaystyle = 2 \sum_{n=1}^{\infty} \int^{\infty}_{0} \Big( \frac{1}{t+n} - \frac{t+n}{(t+n)^{2}+1} \Big) \ dt$

$\displaystyle = \sum_{n=1}^{\infty} \ln \Big( \frac{(t+n)^{2}}{(t+n)^{2}+1} \Big) \Big|^{\infty}_{0}$

$\displaystyle = \sum^{\infty}_{n=1} \ln \Big(1 + \frac{1}{n^{2}} \Big)$

$\displaystyle \ln \Big(\frac{ \sinh \pi a}{\pi a} \Big) = \sum^{\infty}_{n=1} \ln \Big( 1 + \frac{a^{2}}{n^{2}} \Big)$

so $\displaystyle \sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} \zeta (2n) = \ln \Big( \frac{\sinh \pi}{\pi} \Big)$
• Jan 30th 2011, 04:26 PM
chiph588@
Quote:

Originally Posted by Random Variable
$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \ \zeta(2n)$

We know $\displaystyle \frac{\sin\pi x}{\pi x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2}\right) \implies \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n}$, for $|x|<1$.

Thus $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\zeta(2n) = \log(\sin\pi i)-\log(\pi i) = \log(i\sinh\pi)-\log(\pi i) = \log\left(\frac{\sinh\pi}\pi\right)$.

Notice we don't need to consider a which branch we choose here, since things cancel!
• Jan 31st 2011, 12:24 PM
Drexel28
Quote:

Originally Posted by chiph588@
We know $\displaystyle \frac{\sin\pi x}{\pi x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2}\right) \implies \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n}$, for $|x|<1$.

Thus $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\zeta(2n) = \log(\sin\pi i)-\log(\pi i) = \log(i\sinh\pi)-\log(\pi i) = \log\left(\frac{\sinh\pi}\pi\right)$.

Notice we don't need to consider a which branch we choose here, since things cancel!

It's actually true for $-1\leqslant x<1$ which is what enabled you to do this. The proof is slightly less clear than the one you did. You need to consider exactly what you said and then apply Abel's Limit Theorem.