Question

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

Answer 1

Split the summand into partial fractions,

`\frac1{n(2n+1)} = \frac1n - \frac2{2n+1}`

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

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

from which we have

`x \cot(x) = \displaystyle \sum_(n=0)^\infty ((-1)^n B_(2n))/((2n)!) \left(4x^2\right)^n`

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

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

Let

`f(x) = \displaystyle \sum_(n=1)^\infty ((-1)^n B_(2n))/(n (2n)!) x^n`

Differentiating both sides gives

`f'(x) = \displaystyle \sum_(n=1)^\infty ((-1)^n B_(2n))/((2n)!) x^(n-1)`

so that

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

By the fundamental theorem of calculus,

`f'(x) = \frac1{2\sqrt x} \cot\left(\frac{\sqrt x}2\right) - \frac1x`

`\implies f(x) = f(0) + 2 \ln\left(\sin\left(\frac{\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 ((-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 ((-1)^n B_(2n))/((2n)!) (x^2)^n`

We then have

`g'(x) = \frac x2 \cot\left(\frac 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

`(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 (\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 ((-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 (\zeta(2n))/(n(2n+1)4^(2n)) = \boxed{\ln(\pi) - \ln(2) + \frac{2G}\pi - 1}`