Question

$\rm \sum_{n = 1}^ \infty \frac{ \zeta(2n)}{ n(2n + 1) {4}^{2n} }$​

Answer 1

Split the summand into partial fractions,

$\dfrac1{n(2n+1)} = \dfrac1n - \dfrac2{2n+1}$

At even integers, the Riemann zeta function has the equivalent form

$\zeta(2n) = \dfrac{(-1)^{n+1} B_{2n} 2^{2n-1} \pi^{2n}}{(2n)!}$

where $B_n$ denotes the n-th Bernoulli number.

Now the sum we want is

$\displaystyle \sum_{n=1}^\infty \frac{\zeta(2n)}{n(2n+1)4^{2n}} = -\frac12 \sum_{n=1}^\infty \frac{(-1)^n B_{2n}}{n (2n)!} \left(\frac{\pi^2}4\right)^n + \sum_{n=1}^\infty \frac{(-1)^n B_{2n}}{(2n+1) (2n)!} \left(\frac{\pi^2}4\right)^n$

Recall the series expansion for cot(x), valid for 0 < x < π :

$\cot(x) = \displaystyle \sum_{n=0}^\infty \frac{(-1)^n B_{2n} 2^{2n} x^{2n-1}}{(2n)!}$

from which we have

$x \cot(x) = \displaystyle \sum_{n=0}^\infty \frac{(-1)^n B_{2n}}{(2n)!} \left(4x^2\right)^n$

and letting $x \to \frac{\sqrt x}2$, we get

$-\dfrac{\sqrt x}2 \cot\left(\dfrac{\sqrt x}2\right) = \displaystyle \sum_{n=0}^\infty \frac{(-1)^{n+1} B_{2n}}{(2n)!} x^n$

Let

$f(x) = \displaystyle \sum_{n=1}^\infty \frac{(-1)^n B_{2n}}{n (2n)!} x^n$

Differentiating both sides gives

$f'(x) = \displaystyle \sum_{n=1}^\infty \frac{(-1)^n B_{2n}}{(2n)!} x^{n-1}$

so that

$x f'(x) = \dfrac{\sqrt x}2 \cot\left(\dfrac{\sqrt x}2\right) - 1$

By the fundamental theorem of calculus,

$f'(x) = \dfrac1{2\sqrt x} \cot\left(\dfrac{\sqrt x}2\right) - \dfrac1x$

$\implies f(x) = f(0) + 2 \ln\left(\sin\left(\dfrac{\sqrt x}2\right)\right) - \ln(x)$

We observe that as x approaches 0, the series vanishes, so we must have

$\displaystyle f(0) = \lim_{x\to0^+} 2\ln\left(\sin\left(\frac{\sqrt x}2\right)\right) - \ln(x) = -2\ln(2)$

Similarly, let

$g(x) = \displaystyle \sum_{n=1}^\infty \frac{(-1)^n B_{2n}}{(2n+1) (2n)!} x^{2n+1}$

with g(0) = 0, and differentiate to get

$g'(x) = \displaystyle \sum_{n=1}^\infty \frac{(-1)^n B_{2n}}{(2n)!} (x^2)^n$

We then have

$g'(x) = \dfrac x2 \cot\left(\dfrac x2\right) - 1$

and by the fundamental theorem of calculus,

$g(x) = \displaystyle \int_0^x \frac t2 \cot\left(\frac t2\right) \, dt - x$

or

$\dfrac{g(\sqrt x)}{\sqrt x} = \displaystyle \frac1{\sqrt x} \int_0^{\sqrt x} \frac t2 \cot\left(\frac t2\right) \, dt - 1$

The sum we want is then

$\displaystyle \sum_{n=1}^\infty \frac{\zeta(2n)}{n(2n+1)4^{2n}} = -\frac12 f\left(\frac{\pi^2}4\right) + \frac1{\sqrt{\frac{\pi^2}4}} g\left(\sqrt{\frac{\pi^2}4}\right) = -\frac12 f\left(\frac{\pi^2}4\right) + \frac2\pi g\left(\frac\pi2\right)$

Now we turn our attention to

$\displaystyle \int_0^{\frac\pi2} \frac x2 \cot\left(\frac x2\right) \, dx = 2 \int_0^{\frac\pi4} x \cot(x) \, dx$

Integrating by parts gives

$\displaystyle \int_0^{\frac\pi4} x \cot(x) \, dx = -\frac\pi8 \ln(2) - \int_0^{\frac\pi4} \ln(\sin(x)) \, dx$

Using the identity from your earlier question [26989784],

$\displaystyle -\int_0^{\frac\pi4} \ln(\sin(x)) \, dx = \frac\pi8 \ln(2) + \frac12 \sum_{k=1}^\infty \frac1{k^2} \sin\left(\frac{k\pi}2\right) \\\\ = \frac\pi8 \ln(2) + \frac12 \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2} \\\\ = \frac\pi8 \ln(2) + \frac G2$

where G is Catalan's constant.

So, it would seem

$\displaystyle \sum_{n=1}^\infty \frac{\zeta(2n)}{n(2n+1)4^{2n}} = \boxed{\ln(\pi) - \ln(2) + \frac{2G}\pi - 1}$