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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?

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  • $\begingroup$ what is the question? $\endgroup$ – tired Jan 3 '17 at 15:22
  • $\begingroup$ @tired I believe it is $(0)$. $\endgroup$ – Simply Beautiful Art Jan 3 '17 at 15:24
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    $\begingroup$ If (3) and (4) are "known", what is the question? $\endgroup$ – Did Jan 3 '17 at 15:32
  • $\begingroup$ ((OP: It seems that firing "questions" at such a rapid rate does not help. This is not a race, you know?)) $\endgroup$ – Did Jan 3 '17 at 15:34
  • $\begingroup$ Look up and you see it. Prove that: $\endgroup$ – gymbvghjkgkjkhgfkl Jan 3 '17 at 16:16
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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.

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  • $\begingroup$ Thank you @Marco Cantarini. $\endgroup$ – gymbvghjkgkjkhgfkl Jan 4 '17 at 12:30
  • $\begingroup$ @X-men You're welcome :) $\endgroup$ – Marco Cantarini Jan 4 '17 at 12:37
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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

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

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

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