Dtjytyk

Is there a way to generate uniformly distributed points on a sphere from a fixed amount of random real numbers per point?

One-dimensional Japanese puzzle

Are spiders unable to hurt humans, especially very small spiders?

Are there continuous functions who are the same in an interval but differ in at least one other point?

Can I visit the Trinity College (Cambridge) library and see some of their rare books

Word for: a synonym with a positive connotation?

Do I have Disadvantage attacking with an off-hand weapon?

Simulating Exploding Dice

Match Roman Numerals

What aspect of planet Earth must be changed to prevent the industrial revolution?

"is" operation returns false even though two objects have same id

Keeping a retro style to sci-fi spaceships?

Can the Right Ascension and Argument of Perigee of a spacecraft's orbit keep varying by themselves with time?

How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time

How did passengers keep warm on sail ships?

Windows 10: How to Lock (not sleep) laptop on lid close?

Didn't get enough time to take a Coding Test - what to do now?

How to handle characters who are more educated than the author?

1960s short story making fun of James Bond-style spy fiction

Does Parliament hold absolute power in the UK?

Example of compact Riemannian manifold with only one geodesic.

Is there a writing software that you can sort scenes like slides in PowerPoint?

Do working physicists consider Newtonian mechanics to be "falsified"?

Drawing vertical/oblique lines in Metrical tree (tikz-qtree, tipa)

Prove $sum_{n=1}^infty frac{zeta(2n+1) - 1}{(2n+1) 2^{2n}} = 1 + ln(2) - ln(3) - gamma$

3

1

$begingroup$

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

$$sum_{n=1}^infty frac{zeta(2n+1) - 1}{(2n+1) 2^{2n}} = 1 + ln(2) - ln(3) - gamma$$

I know that $$sum_{k=1}^infty frac{zeta(2k) - 1}{k} = ln(2)$$

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.

$$sum_{n=1}^infty frac{zeta(2n+1) - 1}{(2n+1) 2^{2n}} = 1 + ln(2) - ln(3) - gamma$$

I know that $$sum_{k=1}^infty frac{zeta(2k) - 1}{k} = ln(2)$$

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.

$$sum_{n=1}^infty frac{zeta(2n+1) - 1}{(2n+1) 2^{2n}} = 1 + ln(2) - ln(3) - gamma$$

I know that $$sum_{k=1}^infty frac{zeta(2k) - 1}{k} = ln(2)$$

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

$$zeta(2n+1)-1=sum_{k=2}^infty frac{1}{k^{2n+1}}tag1$$

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

$$f(x)=sum_{n=1}^infty frac{zeta(2n+1)-1}{2n+1},x^{2n+1}tag2$$

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

$$begin{align}
f(x)&=int_0^x sum_{n=1}^infty sum_{k=2}^infty frac{t^{2n}}{k^{2n+1}},dt\\
&=int_0^x sum_{k=2}^infty frac1k sum_{n=1}^infty left(frac{t^2}{k^2}right)^n ,dt \\
&=int_0^x sum_{k=2}^infty frac1k frac{t^2}{k^2-t^2},dt\\
&=int_0^x left(frac{t^2}{t^2-1}+sum_{k=1}^infty frac1kfrac{t^2}{k^2-t^2}right),dt\\
&=int_0^x frac{t^2}{t^2-1},dt-frac12int_0^x sum_{k=1}^inftyleft(frac1k-frac1{k+t}right),dt-frac12int_0^x sum_{k=1}^inftyleft(frac1k-frac1{k-t}right),dt\\
&=int_0^x frac{t^2}{t^2-1},dt-frac12 int_0^x left(2gamma +psi(t+1)+psi(1-t)right),dt\\
&=x+logleft(frac{1-x}{1+x}right)-gamma x -frac12 logleft(frac{Gamma(1+x)}{Gamma(1-x)}right)
end{align}$$

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

$$sum_{n=1}^infty frac{zeta(2n+1)-1}{(2n+1)2^{2n}}=1-log(3)-gamma+log(2)$$

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 | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3158109%2fprove-sum-n-1-infty-frac-zeta2n1-12n1-22n-1-ln2-l%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

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

$$zeta(2n+1)-1=sum_{k=2}^infty frac{1}{k^{2n+1}}tag1$$

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

$$f(x)=sum_{n=1}^infty frac{zeta(2n+1)-1}{2n+1},x^{2n+1}tag2$$

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

$$begin{align}
f(x)&=int_0^x sum_{n=1}^infty sum_{k=2}^infty frac{t^{2n}}{k^{2n+1}},dt\\
&=int_0^x sum_{k=2}^infty frac1k sum_{n=1}^infty left(frac{t^2}{k^2}right)^n ,dt \\
&=int_0^x sum_{k=2}^infty frac1k frac{t^2}{k^2-t^2},dt\\
&=int_0^x left(frac{t^2}{t^2-1}+sum_{k=1}^infty frac1kfrac{t^2}{k^2-t^2}right),dt\\
&=int_0^x frac{t^2}{t^2-1},dt-frac12int_0^x sum_{k=1}^inftyleft(frac1k-frac1{k+t}right),dt-frac12int_0^x sum_{k=1}^inftyleft(frac1k-frac1{k-t}right),dt\\
&=int_0^x frac{t^2}{t^2-1},dt-frac12 int_0^x left(2gamma +psi(t+1)+psi(1-t)right),dt\\
&=x+logleft(frac{1-x}{1+x}right)-gamma x -frac12 logleft(frac{Gamma(1+x)}{Gamma(1-x)}right)
end{align}$$

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

$$sum_{n=1}^infty frac{zeta(2n+1)-1}{(2n+1)2^{2n}}=1-log(3)-gamma+log(2)$$

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

$$zeta(2n+1)-1=sum_{k=2}^infty frac{1}{k^{2n+1}}tag1$$

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

$$f(x)=sum_{n=1}^infty frac{zeta(2n+1)-1}{2n+1},x^{2n+1}tag2$$

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

$$begin{align}
f(x)&=int_0^x sum_{n=1}^infty sum_{k=2}^infty frac{t^{2n}}{k^{2n+1}},dt\\
&=int_0^x sum_{k=2}^infty frac1k sum_{n=1}^infty left(frac{t^2}{k^2}right)^n ,dt \\
&=int_0^x sum_{k=2}^infty frac1k frac{t^2}{k^2-t^2},dt\\
&=int_0^x left(frac{t^2}{t^2-1}+sum_{k=1}^infty frac1kfrac{t^2}{k^2-t^2}right),dt\\
&=int_0^x frac{t^2}{t^2-1},dt-frac12int_0^x sum_{k=1}^inftyleft(frac1k-frac1{k+t}right),dt-frac12int_0^x sum_{k=1}^inftyleft(frac1k-frac1{k-t}right),dt\\
&=int_0^x frac{t^2}{t^2-1},dt-frac12 int_0^x left(2gamma +psi(t+1)+psi(1-t)right),dt\\
&=x+logleft(frac{1-x}{1+x}right)-gamma x -frac12 logleft(frac{Gamma(1+x)}{Gamma(1-x)}right)
end{align}$$

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

$$sum_{n=1}^infty frac{zeta(2n+1)-1}{(2n+1)2^{2n}}=1-log(3)-gamma+log(2)$$

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

$$zeta(2n+1)-1=sum_{k=2}^infty frac{1}{k^{2n+1}}tag1$$

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

$$f(x)=sum_{n=1}^infty frac{zeta(2n+1)-1}{2n+1},x^{2n+1}tag2$$

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

$$begin{align}
f(x)&=int_0^x sum_{n=1}^infty sum_{k=2}^infty frac{t^{2n}}{k^{2n+1}},dt\\
&=int_0^x sum_{k=2}^infty frac1k sum_{n=1}^infty left(frac{t^2}{k^2}right)^n ,dt \\
&=int_0^x sum_{k=2}^infty frac1k frac{t^2}{k^2-t^2},dt\\
&=int_0^x left(frac{t^2}{t^2-1}+sum_{k=1}^infty frac1kfrac{t^2}{k^2-t^2}right),dt\\
&=int_0^x frac{t^2}{t^2-1},dt-frac12int_0^x sum_{k=1}^inftyleft(frac1k-frac1{k+t}right),dt-frac12int_0^x sum_{k=1}^inftyleft(frac1k-frac1{k-t}right),dt\\
&=int_0^x frac{t^2}{t^2-1},dt-frac12 int_0^x left(2gamma +psi(t+1)+psi(1-t)right),dt\\
&=x+logleft(frac{1-x}{1+x}right)-gamma x -frac12 logleft(frac{Gamma(1+x)}{Gamma(1-x)}right)
end{align}$$

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

$$sum_{n=1}^infty frac{zeta(2n+1)-1}{(2n+1)2^{2n}}=1-log(3)-gamma+log(2)$$

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