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Prove $sum_n=1^infty fraczeta(2n+1) - 1(2n+1) 2^2n = 1 + ln(2) - ln(3) - gamma$

Euler-Mascheroni constant: understanding why $lim_mrightarrow infty sum_n=1^m (ln (1 + frac1n)-frac1n+1)= 1 - gamma$How to show $gamma =sum_m=2^infty(-1)^mfraczeta (m)m$?Why is $1 + frac12 + frac13 + … + frac1n approx ln(n) + gamma$?Finding $sum_k=1^infty left[frac12k-log left(1+frac12kright)right]$Proving $lim_ntoinftysum_k=n^inftyBig(frac1nleftlfloorfrac knrightrfloor-frac1kBig)=gamma$How to prove this limit about $gamma=lim_Nto infty left(sum_n=1^Nfrac1n-ln Nright)$How to show that $gamma=limlimits_n to inftyleft(sumlimits_k=1^nzeta(2k)over kright)-ln(2n)$What is the series representation of $frac1gamma$?Prove that $-2log(2) = -2 + sum_n=1^inftyfrac1n(2n+1)$Divergence of $sum_n=1^inftyfracmu(n)sqrtncosleft(n^2 pi gammaright)$, where $gamma$ is the Euler-Mascheroni constant

3

1

$begingroup$

I've found the following series on the Wikipediapage of the Euler-Mascheroni constant and I want to prove it.

$$sum_n=1^infty fraczeta(2n+1) - 1(2n+1) 2^2n = 1 + ln(2) - ln(3) - gamma$$

I know that $$sum_k=1^infty fraczeta(2k) - 1k = ln(2)$$

Any help would be appreciated. Thanks in advance.

summation logarithms zeta-functions eulers-constant

share|cite|improve this question

edited Mar 23 at 9:33

Andrews

1,2812423

asked Mar 22 at 12:58

Kevin2806Kevin2806

214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
  $endgroup$
  – Mark Viola
  Apr 3 at 3:32
  

  
  
  
  

add a comment | 

3

1

$begingroup$

I've found the following series on the Wikipediapage of the Euler-Mascheroni constant and I want to prove it.

$$sum_n=1^infty fraczeta(2n+1) - 1(2n+1) 2^2n = 1 + ln(2) - ln(3) - gamma$$

I know that $$sum_k=1^infty fraczeta(2k) - 1k = ln(2)$$

Any help would be appreciated. Thanks in advance.

summation logarithms zeta-functions eulers-constant

share|cite|improve this question

edited Mar 23 at 9:33

Andrews

1,2812423

asked Mar 22 at 12:58

Kevin2806Kevin2806

214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
  $endgroup$
  – Mark Viola
  Apr 3 at 3:32
  

  
  
  
  

add a comment | 

$begingroup$

I've found the following series on the Wikipediapage of the Euler-Mascheroni constant and I want to prove it.

$$sum_n=1^infty fraczeta(2n+1) - 1(2n+1) 2^2n = 1 + ln(2) - ln(3) - gamma$$

I know that $$sum_k=1^infty fraczeta(2k) - 1k = ln(2)$$

Any help would be appreciated. Thanks in advance.

summation logarithms zeta-functions eulers-constant

share|cite|improve this question

edited Mar 23 at 9:33

Andrews

1,2812423

asked Mar 22 at 12:58

Kevin2806Kevin2806

214

$endgroup$

share|cite|improve this question

edited Mar 23 at 9:33

Andrews

1,2812423

asked Mar 22 at 12:58

Kevin2806Kevin2806

214

share|cite|improve this question

edited Mar 23 at 9:33

Andrews

1,2812423

asked Mar 22 at 12:58

Kevin2806Kevin2806

214

edited Mar 23 at 9:33

Andrews

1,2812423

edited Mar 23 at 9:33

Andrews

1,2812423

asked Mar 22 at 12:58

Kevin2806Kevin2806

214

asked Mar 22 at 12:58

Kevin2806Kevin2806

214

1 Answer
1

active

oldest

votes

4

$begingroup$

First, we use the familiar series representation of the Riemann Zeta function to write

$$zeta(2n+1)-1=sum_k=2^infty frac1k^2n+1tag1$$

Next, define the function $f(x)$ as

$$f(x)=sum_n=1^infty fraczeta(2n+1)-12n+1,x^2n+1tag2$$

Using $(1)$ in $(2)$ reveals

$$beginalign
f(x)&=int_0^x sum_n=1^infty sum_k=2^infty fract^2nk^2n+1,dt\\
&=int_0^x sum_k=2^infty frac1k sum_n=1^infty left(fract^2k^2right)^n ,dt \\
&=int_0^x sum_k=2^infty frac1k fract^2k^2-t^2,dt\\
&=int_0^x left(fract^2t^2-1+sum_k=1^infty frac1kfract^2k^2-t^2right),dt\\
&=int_0^x fract^2t^2-1,dt-frac12int_0^x sum_k=1^inftyleft(frac1k-frac1k+tright),dt-frac12int_0^x sum_k=1^inftyleft(frac1k-frac1k-tright),dt\\
&=int_0^x fract^2t^2-1,dt-frac12 int_0^x left(2gamma +psi(t+1)+psi(1-t)right),dt\\
&=x+logleft(frac1-x1+xright)-gamma x -frac12 logleft(fracGamma(1+x)Gamma(1-x)right)
endalign$$

Setting $t=1/2$ and multiplying by $2$ reveals

$$sum_n=1^infty fraczeta(2n+1)-1(2n+1)2^2n=1-log(3)-gamma+log(2)$$

as was to be shown!

share|cite|improve this answer

answered Mar 22 at 14:50

Mark ViolaMark Viola

134k1278177

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) Beautiful!
  $endgroup$
  – mrtaurho
  Mar 22 at 17:21
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @mrtaurho Thank you! Very much appreciative.
  $endgroup$
  – Mark Viola
  Mar 22 at 17:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kevin2806 Please let me know how I can improve my answer. I really want to give you the best answer I can.
  $endgroup$
  – Mark Viola
  Mar 25 at 17:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And feel free to up vote and accept an answer as you see fit.
  $endgroup$
  – Mark Viola
  Mar 25 at 17:46
  

  
  
  
  

add a comment | 

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4

$begingroup$

First, we use the familiar series representation of the Riemann Zeta function to write

$$zeta(2n+1)-1=sum_k=2^infty frac1k^2n+1tag1$$

Next, define the function $f(x)$ as

$$f(x)=sum_n=1^infty fraczeta(2n+1)-12n+1,x^2n+1tag2$$

Using $(1)$ in $(2)$ reveals

$$beginalign
f(x)&=int_0^x sum_n=1^infty sum_k=2^infty fract^2nk^2n+1,dt\\
&=int_0^x sum_k=2^infty frac1k sum_n=1^infty left(fract^2k^2right)^n ,dt \\
&=int_0^x sum_k=2^infty frac1k fract^2k^2-t^2,dt\\
&=int_0^x left(fract^2t^2-1+sum_k=1^infty frac1kfract^2k^2-t^2right),dt\\
&=int_0^x fract^2t^2-1,dt-frac12int_0^x sum_k=1^inftyleft(frac1k-frac1k+tright),dt-frac12int_0^x sum_k=1^inftyleft(frac1k-frac1k-tright),dt\\
&=int_0^x fract^2t^2-1,dt-frac12 int_0^x left(2gamma +psi(t+1)+psi(1-t)right),dt\\
&=x+logleft(frac1-x1+xright)-gamma x -frac12 logleft(fracGamma(1+x)Gamma(1-x)right)
endalign$$

Setting $t=1/2$ and multiplying by $2$ reveals

$$sum_n=1^infty fraczeta(2n+1)-1(2n+1)2^2n=1-log(3)-gamma+log(2)$$

as was to be shown!

share|cite|improve this answer

answered Mar 22 at 14:50

Mark ViolaMark Viola

134k1278177

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) Beautiful!
  $endgroup$
  – mrtaurho
  Mar 22 at 17:21
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @mrtaurho Thank you! Very much appreciative.
  $endgroup$
  – Mark Viola
  Mar 22 at 17:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kevin2806 Please let me know how I can improve my answer. I really want to give you the best answer I can.
  $endgroup$
  – Mark Viola
  Mar 25 at 17:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And feel free to up vote and accept an answer as you see fit.
  $endgroup$
  – Mark Viola
  Mar 25 at 17:46
  

  
  
  
  

add a comment | 

4

$begingroup$

First, we use the familiar series representation of the Riemann Zeta function to write

$$zeta(2n+1)-1=sum_k=2^infty frac1k^2n+1tag1$$

Next, define the function $f(x)$ as

$$f(x)=sum_n=1^infty fraczeta(2n+1)-12n+1,x^2n+1tag2$$

Using $(1)$ in $(2)$ reveals

$$beginalign
f(x)&=int_0^x sum_n=1^infty sum_k=2^infty fract^2nk^2n+1,dt\\
&=int_0^x sum_k=2^infty frac1k sum_n=1^infty left(fract^2k^2right)^n ,dt \\
&=int_0^x sum_k=2^infty frac1k fract^2k^2-t^2,dt\\
&=int_0^x left(fract^2t^2-1+sum_k=1^infty frac1kfract^2k^2-t^2right),dt\\
&=int_0^x fract^2t^2-1,dt-frac12int_0^x sum_k=1^inftyleft(frac1k-frac1k+tright),dt-frac12int_0^x sum_k=1^inftyleft(frac1k-frac1k-tright),dt\\
&=int_0^x fract^2t^2-1,dt-frac12 int_0^x left(2gamma +psi(t+1)+psi(1-t)right),dt\\
&=x+logleft(frac1-x1+xright)-gamma x -frac12 logleft(fracGamma(1+x)Gamma(1-x)right)
endalign$$

Setting $t=1/2$ and multiplying by $2$ reveals

$$sum_n=1^infty fraczeta(2n+1)-1(2n+1)2^2n=1-log(3)-gamma+log(2)$$

as was to be shown!

share|cite|improve this answer

answered Mar 22 at 14:50

Mark ViolaMark Viola

134k1278177

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) Beautiful!
  $endgroup$
  – mrtaurho
  Mar 22 at 17:21
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @mrtaurho Thank you! Very much appreciative.
  $endgroup$
  – Mark Viola
  Mar 22 at 17:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kevin2806 Please let me know how I can improve my answer. I really want to give you the best answer I can.
  $endgroup$
  – Mark Viola
  Mar 25 at 17:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And feel free to up vote and accept an answer as you see fit.
  $endgroup$
  – Mark Viola
  Mar 25 at 17:46
  

  
  
  
  

add a comment | 

$begingroup$

First, we use the familiar series representation of the Riemann Zeta function to write

$$zeta(2n+1)-1=sum_k=2^infty frac1k^2n+1tag1$$

Next, define the function $f(x)$ as

$$f(x)=sum_n=1^infty fraczeta(2n+1)-12n+1,x^2n+1tag2$$

Using $(1)$ in $(2)$ reveals

$$beginalign
f(x)&=int_0^x sum_n=1^infty sum_k=2^infty fract^2nk^2n+1,dt\\
&=int_0^x sum_k=2^infty frac1k sum_n=1^infty left(fract^2k^2right)^n ,dt \\
&=int_0^x sum_k=2^infty frac1k fract^2k^2-t^2,dt\\
&=int_0^x left(fract^2t^2-1+sum_k=1^infty frac1kfract^2k^2-t^2right),dt\\
&=int_0^x fract^2t^2-1,dt-frac12int_0^x sum_k=1^inftyleft(frac1k-frac1k+tright),dt-frac12int_0^x sum_k=1^inftyleft(frac1k-frac1k-tright),dt\\
&=int_0^x fract^2t^2-1,dt-frac12 int_0^x left(2gamma +psi(t+1)+psi(1-t)right),dt\\
&=x+logleft(frac1-x1+xright)-gamma x -frac12 logleft(fracGamma(1+x)Gamma(1-x)right)
endalign$$

Setting $t=1/2$ and multiplying by $2$ reveals

$$sum_n=1^infty fraczeta(2n+1)-1(2n+1)2^2n=1-log(3)-gamma+log(2)$$

as was to be shown!

share|cite|improve this answer

answered Mar 22 at 14:50

Mark ViolaMark Viola

134k1278177

$endgroup$

share|cite|improve this answer

answered Mar 22 at 14:50

Mark ViolaMark Viola

134k1278177

answered Mar 22 at 14:50

Mark ViolaMark Viola

134k1278177

answered Mar 22 at 14:50

Mark ViolaMark Viola

134k1278177