Prove that : $\sum_{n=1}^{\infty}\left({1\over n}-{2\zeta(2n)\over 2n+1}\right)=\ln{\pi}-1$

Question

Prove that

$$\sum_{n=1}^{\infty}\left({1\over n}-{2\zeta(2n)\over 2n+1}\right)=\ln{\pi}-1\tag0$$

My try:

$$\sum_{n=1}^{\infty}\left({1\over n}-{1\over n+1}\right)=1\tag1$$

From wolfram:(126)

$$\sum_{n=1}^{\infty}{\zeta(2n)\over n(2n+1)}=\ln{2\pi}-1\tag2$$

Rewrite:

$$\sum_{n=1}^{\infty}\left({\zeta(2n)\over 2n}-{\zeta(2n)\over 2n+1}\right)={1\over 2}\left(\ln{2\pi}-1\right)\tag3$$

$2\times(3)-(0)$:

$$\sum_{n=1}^{\infty}\left({\zeta(2n)-1\over n}\right)=\ln{2}\tag4$$

$(4)$ it is a well known proven identity from wikipedia(infinite series)

If I got to $(4)$ and it is know on that (4) is correct then it is imply that my original question must be correct?

Answer 1

Note that $$\sum_{n\geq1}\left(\frac{1}{n}-\frac{2\zeta\left(2n\right)}{2n+1}\right)=\sum_{n\geq1}\left(\frac{2n+1-2n\zeta\left(2n\right)}{n\left(2n+1\right)}\right) $$ $$=2\sum_{n\geq1}\left(\frac{1-\zeta\left(2n\right)}{2n+1}\right)+\sum_{n\geq1}\left(\frac{1}{n\left(2n+1\right)}\right)=S_{1}+S_{2}, $$ say. So let us analyze $S_{1} $. From the generating function $$\sum_{n\geq1}\zeta\left(2n\right)x^{2n}=\frac{1-\pi x\cot\left(\pi x\right)}{2},\,\left|x\right|<1 $$ we get $$\sum_{n\geq1}\frac{\zeta\left(2n\right)x^{2n+1}}{2n+1}=\frac{1}{2}\left(x-\int_{0}^{x}\pi y\cot\left(\pi y\right)dy\right)\tag{1} $$ and the integral in the RHS of $(1)$ is, integrating by parts, $$\int_{0}^{x}\pi y\cot\left(\pi y\right)dy=x\log\left(\sin\left(\pi x\right)\right)-\frac{1}{\pi}\int_{0}^{\pi x}\log\left(\sin\left(u\right)\right)du $$ $$=x\log\left(\sin\left(\pi x\right)\right)+\frac{1}{2\pi}\textrm{Cl}_{2}\left(2\pi x\right)+x\log\left(2\right) $$ where $\textrm{Cl}_{2}\left(x\right) $ is the Clausen function of order $2$. Now recalling Taylor series of $\textrm{arctanh}(x) $ $$\sum_{n\geq1}\frac{x^{2n+1}}{2n+1}=\textrm{arctanh}(x)-x $$ we get $$\sum_{n\geq1}\frac{\left(1-\zeta\left(2n\right)\right)x^{2n+1}}{2n+1}=\textrm{arctanh}(x)-x-\frac{1}{2}\left(x-x\log\left(\sin\left(\pi x\right)\right)-\frac{1}{2\pi}\textrm{Cl}_{2}\left(2\pi x\right)-x\log\left(2\right)\right)=f(x)$$ and recalling that $$\textrm{Cl}_{2}\left(m\pi\right)=0,\, m\in\mathbb{Z} $$ we get $$\sum_{n\geq1}\frac{\left(1-\zeta\left(2n\right)\right)}{2n+1}=\lim_{x\rightarrow1} f(x)=-\frac{3}{2}+\frac{\log\left(4\pi\right)}{2}. $$ Now we consider $S_{2} $. From the Taylor series of $\log\left(1-x\right) $ we observe that $$\sum_{n\geq1}\frac{x^{2n+1}}{n\left(2n+1\right)}=-\int_{0}^{x}\log\left(1-y^{2}\right)dy=-\int_{0}^{x}\log\left(1-y\right)dy-\int_{0}^{x}\log\left(1+y\right)dy $$ $$=-\left(1-x\right)\log\left(1-x\right)-\left(x+1\right)\log\left(x+1\right)+2x $$ hence $$\sum_{n\geq1}\frac{1}{n\left(2n+1\right)}=\lim_{x\rightarrow1}\left(-\left(1-x\right)\log\left(1-x\right)-\left(x+1\right)\log\left(x+1\right)+2x\right) $$ $$=2-\log\left(4\right) $$ so finally $$\sum_{n\geq1}\left(\frac{1}{n}-\frac{2\zeta\left(2n\right)}{2n+1}\right)=\color{red}{\log\left(\pi\right)-1}$$ as wanted.

Answer 2

Using the generalisation $Q_m(x)$ of the Gamma function in

https://www.fernuni-hagen.de/analysis/download/bachelorarbeit_aschauer.pdf

with $\enspace\displaystyle \ln Q_m(x)=\frac{(-x)^{m+1}}{m+1}\gamma +\sum\limits_{n=2}^\infty\frac{(-x)^{m+n}}{m+n}\zeta(n)\enspace$ on page $13$, $(4.2)$, $Q_m:=Q_m(1),$

and $\enspace\displaystyle Q_m^*(x):=(1+x)Q_m(x)\enspace$ on page $24$, $(4.8)$ and $(4.9)$,

and with $\enspace\displaystyle Q_1^*(1)=2Q_1(1)=2\frac{\sqrt{2\pi}}{e}\enspace$ and $\enspace\displaystyle Q_1^*(-1)=\frac{1}{eQ_1(1)}=\frac{1}{\sqrt{2\pi}}\enspace$ we get

$\enspace\displaystyle\sum\limits_{n=1}^\infty(\frac{1}{n}-\frac{2\zeta(2n)}{2n+1})=-2\sum\limits_{n=1}^\infty(\frac{\zeta(2n)}{2n+1}-\frac{1}{2n+1})+2\sum\limits_{n=1}^\infty(\frac{1}{2n}-\frac{1}{2n+1})=$

$\enspace\displaystyle =-2+\ln Q_1^*(1)-\ln Q_1^*(-1)+2(1-\ln 2)=\ln\pi -1$

Answer 3

We have to prove the relation

$$s:=\sum _{n=1}^{\infty } \left(\frac{1}{n}-\frac{2 \zeta (2 n)}{2 n+1}\right)=\log(\pi)-1\tag{1}$$

Instead of the sum $s$ we consider the generating sum ($0<z<1$)

$$g(z)=\sum _{n=1}^{\infty } z^{2n}\left(\frac{1}{n}-\frac{2 \zeta (2 n)}{2 n+1}\right)=g_1(z)+g_2(z)\tag{2}$$

so that we have $s=\underset{z\to 1_{-}}{\text{lim}}(g(z))$.

Now $$g_1(z) = -\log(1-z^2)\tag{3}$$

and

$$\begin{align} g_2(z) & := -2 \sum _{n=1}^{\infty } \frac{ \zeta (2 n)z^{2 n}}{2 n+1}\\ & = -2 \sum _{n=1}^{\infty } \zeta (2 n) \frac{1}{z}\int_0^z t^{2 n} \, dt \\ & = -\frac{2}{z} \int_0^z \left(\sum _{n=1}^{\infty } \zeta (2 n) t^{2 n}\right) \, dt\\ & = -\frac{2}{z}\sum _{k=1}^{\infty } \int_0^z \left(\sum _{n=1}^{\infty } \frac{t^{2 n}}{k^{2 n}}\right) \, dt \\ & = -\frac{2}{z} \sum _{k=1}^{\infty } \int_0^z \left(-\frac{t^2}{t^2-k^2} \right)\, dt\\ & = \frac{2}{z}\sum _{k=1}^{\infty } \left(z-\frac{1}{2} k \log \left(\frac{k+z}{k-z}\right)\right) \end{align} \tag{4}$$

In order to do the limit $z\to 1$ we consider $g_1$ and the first term of the sum of $g_2$ together plus the remaining sum:

$$s=\underset{z\to 1}{\text{lim}}(g(z))= \underset{z\to 1}{\text{lim}}(-\log(1-z^2) + \left(z-\frac{1}{2} \log \left(\frac{1+z}{1-z}\right)\right))+ s_3\\= 2 (1-\log (2))+s3\tag{5}$$

here $s3=\underset{m\to \infty}{\text{lim}}s_3(m)$ where

$$s_3(m) = 2 \sum _{k=2}^{m } \left(\frac{1}{2} k \log \left(\frac{k+1}{k-1}\right)-1\right)\tag{6}$$

This tough looking sum can luckily be done exactly by elementary operations, i.e. shifting summation indices and summation limits, with the result

$$s_3(m) =2 \log (m!)+2 (m-1)-m \log \left(\frac{1}{m}+1\right)-2 m \log (m)-\log (m)+\log (2)\tag{7}$$

For $m\to \infty$ we have two non-trivial expansions

$$\log \left(\frac{1}{m}+1\right) \simeq \frac{1}{m}-\frac{1}{2 m^2}\tag{8} $$

and the Stirling expansion

$$m!\simeq \left(\frac{\sqrt{\frac{\pi }{2}}}{6 \sqrt{m}}+\sqrt{2 \color{red}\pi } \sqrt{m}\right) e^{m (\log (m)-1)}\tag{9}$$

which is responsible for bringing in the constant $\pi$.

Adding the asymptotic terms and going to the limit $m=\infty$ we obtain

$$s_3 = -3+\log (2)+\log (2 \pi )\tag{10}$$

so that finally

$$s=2 (1-\log (2))-3+\log (2)+\log (2 \pi )=\log(\pi)-1\tag{11}$$

which completes the derivation.