Calculating $\sum_{n=1}^{\infty} \frac{\zeta(2n)}{n \cdot 4^n}$

Question

This is my first post on MSE, so I apologize in advance for any mistakes I may have made.

I was trying to find the value of the sum

$$ S = \sum_{n=1}^{\infty} \frac{\zeta(2n)}{n\cdot 4^n}$$

According to WolframAlpha, this sum evaluates to

$$ S = 0.4515827052894548647....$$

Which WolframAlpha suggests it to be equal to $\ln{\frac{\pi}{2}}$. I have confirmed that it matches exactly up to $20$ decimal places (may be even higher).

So to think about this, I defined a function

$$ S(a) = \sum_{n=1}^{\infty} \frac{\zeta(2n)}{n\cdot a^n}$$

Then differentiating, we have

$$ S'(a) = \sum_{n=1}^{\infty} \frac{\zeta(2n)}{n} \cdot \frac{-n}{a^{n+1}} = \sum_{n=1}^{\infty} -\frac{\zeta(2n)}{a^{n+1}}$$

But I am unable to evaluate $S'(a)$. Can someone give me some hint to find an expression for $S'(a)$?

Many thanks for your efforts!

Answer 1

$$\eqalign{ & S = \sum\limits_{n = 1}^\infty {\frac{{\zeta \left( {2n} \right)}}{{n{4^n}}}} = \sum\limits_{n = 1}^\infty {\frac{1}{{n{4^n}}}} \sum\limits_{k = 1}^\infty {\frac{1}{{{k^{2n}}}}} = \sum\limits_{k = 1}^\infty {\sum\limits_{n = 1}^\infty {\frac{1}{{n{{\left( {4{k^2}} \right)}^n}}}} } \cr & \because \sum\limits_{n = 1}^\infty {\frac{1}{n}} {x^n} = - \ln \left( {1 - x} \right) \Rightarrow \sum\limits_{k = 1}^\infty {\sum\limits_{n = 1}^\infty {\frac{1}{{n{{\left( {4{k^2}} \right)}^n}}}} } = - \sum\limits_{k = 1}^\infty {\ln \left( {1 - \frac{1}{{4{k^2}}}} \right)} \cr & = - \ln \left( {\prod\limits_{k = 1}^\infty {\left( {1 - \frac{1}{{4{k^2}}}} \right)} } \right) = - \ln \left( {\prod\limits_{k = 1}^\infty {\left( {1 - {{\left( {\frac{1}{{2k}}} \right)}^2}} \right)} } \right) \cr & \because \frac{{\sin \left( {\pi z} \right)}}{{\pi z}} = \prod\limits_{k = 1}^\infty {\left( {1 - {{\left( {\frac{z}{k}} \right)}^2}} \right)} ,{\text{ let}}:z = \frac{1}{2} \Rightarrow \frac{{\sin \left( {\frac{\pi }{2}} \right)}}{{\frac{\pi }{2}}} = \prod\limits_{k = 1}^\infty {\left( {1 - {{\left( {\frac{1}{{2k}}} \right)}^2}} \right)} \Leftrightarrow \prod\limits_{k = 1}^\infty {\left( {1 - {{\left( {\frac{1}{{2k}}} \right)}^2}} \right)} = \frac{2}{\pi } \cr & \Rightarrow S = - \ln \left( {\frac{2}{\pi }} \right) = \ln \left( {\frac{\pi }{2}} \right) \cr} $$

Answer 2

To compute the value of the sum

$$S(a) = \sum_{n=1}^{\infty} \frac{\zeta(2n)}{n \cdot a^n}$$

Expand and rearrange to obtain

$$S(a) = \frac{\zeta(2)}{1\cdot a} + \frac{\zeta(4)}{2 \cdot a^2} + \frac{\zeta(6)}{ 3 \cdot a^3}.....$$

Note that

$$\zeta(2n) = \sum_{k=0}^{\infty} \frac{1}{(k+1)^{2n}}$$

Recall the Taylor Series expansion of $\ln(1-x)$

$$\ln(1-x) = -x -\frac{x^2}{2} -\frac{x^3}{3}.....$$

This implies that

$$-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3}.... = \sum_{n=1}^{\infty} \frac{x^n}{n}$$

Thus, using the above two identities, we can rewrite $S(a)$ as:

$$S(a) = \sum_{n=0}^{\infty} -\ln(1-\frac{1}{a(n+1)^2})$$

Note that

$$-\ln(1-\frac{1}{a(n+1)^2}) = \int_0^{\frac{1}{\sqrt{a}}} \frac{2a}{(n+1)^2-a^2} da$$

This means that

$$S(a) = \sum_{n=0}^{\infty} \int_0^{\frac{1}{\sqrt{a}}} \frac{2a}{(n+1)^2-a^2} da$$

Swapping the integral and summation signs, we get:

$$S(a) = \int_0^{\frac{1}{\sqrt{a}}} \sum_{n=0}^{\infty} \frac{2a}{(n+1)^2-a^2} da$$

Also, we have:

$$\sum_{n=0}^{\infty} \frac{2a}{(n+1)^2-a^2} = \frac{1-\pi a \cdot \cot(\pi a)}{a}$$

Thus,

$$S(a) = \int_0^{\frac{1}{\sqrt{a}}} \frac{1-\pi a \cdot \cot(\pi a)}{a} da$$

$$S(a) = \ln(\frac{1}{\sqrt{a}}) - \ln(\sin(\frac{\pi}{\sqrt{a}})) - \ln(\frac{1}{\pi})$$

We finally obtain:

$$S(a) = \ln(\frac{\pi}{ \sqrt{a}} \cdot \csc(\frac{\pi}{ \sqrt{a}}))$$

Substituting $a=4$, we get the desired result:

$$S(4) = \ln(\frac{\pi}{2})$$

P.S. : We can also compute the sum

$$S_1(a) = \sum_{n=1}^{\infty} \frac{\zeta(2n)}{a^{n}}$$

By your earlier observation that:

$$S'(a) = -\frac{1}{a} \cdot \sum_{n=1}^{\infty} \frac{\zeta(2n)}{a^n} = \frac{d}{da}\ln(\frac{\pi}{ \sqrt{a}} \cdot \csc(\frac{\pi}{ \sqrt{a}}))$$

By observing that the sum in LHS is $S_1(a)$ multiplied by $\frac{-1}{a}$, we obtain:

$$S_1(a) = \sum_{n=1}^{\infty} \frac{\zeta(2n)}{a^n} = \frac{1}{2} \cdot \left(\csc(\frac{\pi}{\sqrt{x}}) - \frac{\pi}{\sqrt{x}}\cdot \ln(\frac{\pi}{\sqrt{x}}) \cdot \cot(\frac{\pi}{\sqrt{x}}\right) $$

Answer 3

Another way...

By using the well-known identity $\Gamma(s)\zeta(s)=\int_0^{\infty}\frac{x^{s-1}}{e^x-1}dx$ we have $$\sum_{n=1}^{\infty}\frac{\zeta(2n)}{n4^n}= \sum_{n=1}^{\infty}\frac{ \int_0^{\infty}\frac{x^{2n-1}}{e^x-1}dx}{(2n-1)!n4^n}\\ =\int_0^\infty\frac2{x(e^x-1)}\sum_{n=1}^{\infty}\frac{(\tfrac x2)^{2n}}{(2n)!}dx =\int_0^\infty\frac{2(\cosh(\tfrac x2)-1)}{x(e^x-1)}dx =\int_0^\infty\frac{(e^{\frac x2}-1)^2}{xe^{\frac x2}(e^x-1)}dx =\int_0^\infty\frac{(e^x-1)^2}{xe^x(e^{2x}-1)}dx =\int_0^\infty\frac{e^x-1}{xe^x(e^x+1)}dx\\ \stackrel{t=e^{-x}}{=}\int_0^1\frac{t-1}{\ln t(t+1)}dt\\ =J(0) $$ where $J(0)=\ln(\tfrac\pi 2)$ due to Svyatoslav's answer here.