What's the series of $sum_ngeqslant1 dfraczeta(2n)n2^2n$.

Clash Royale CLAN TAG#URR8PPP

up vote
5
down vote

favorite

2

I know with the formula
$$1-sum_ngeq 12zeta(2n),x^2n=pi xcot(pi x)$$
may I find the following relation used here

$$
sum_ngeqslant1 dfraczeta(2n)n2^2n=colorbluelndfracpi2
$$

hardly, since I have
$$intsum_ngeq 1zeta(2n),x^n-1dx=intleft(dfrac12x^n+1-dfracpi2 dfraccot(pi x)x^nright)dx$$
and after integration set $x=dfrac14$, but it seems so hard.

Any suggestion, thanks in advanced!

integration zeta-functions

share|cite|improve this question

edited Jul 20 at 13:54

asked Jul 20 at 12:57

Nosrati

19.5k41544

add a comment | 

up vote
5
down vote

favorite

2

I know with the formula
$$1-sum_ngeq 12zeta(2n),x^2n=pi xcot(pi x)$$
may I find the following relation used here

$$
sum_ngeqslant1 dfraczeta(2n)n2^2n=colorbluelndfracpi2
$$

hardly, since I have
$$intsum_ngeq 1zeta(2n),x^n-1dx=intleft(dfrac12x^n+1-dfracpi2 dfraccot(pi x)x^nright)dx$$
and after integration set $x=dfrac14$, but it seems so hard.

Any suggestion, thanks in advanced!

integration zeta-functions

share|cite|improve this question

edited Jul 20 at 13:54

asked Jul 20 at 12:57

Nosrati

19.5k41544

add a comment | 

up vote
5
down vote

favorite

2

up vote
5
down vote

favorite

2

2

I know with the formula
$$1-sum_ngeq 12zeta(2n),x^2n=pi xcot(pi x)$$
may I find the following relation used here

$$
sum_ngeqslant1 dfraczeta(2n)n2^2n=colorbluelndfracpi2
$$

hardly, since I have
$$intsum_ngeq 1zeta(2n),x^n-1dx=intleft(dfrac12x^n+1-dfracpi2 dfraccot(pi x)x^nright)dx$$
and after integration set $x=dfrac14$, but it seems so hard.

Any suggestion, thanks in advanced!

integration zeta-functions

share|cite|improve this question

edited Jul 20 at 13:54

asked Jul 20 at 12:57

Nosrati

19.5k41544

I know with the formula
$$1-sum_ngeq 12zeta(2n),x^2n=pi xcot(pi x)$$
may I find the following relation used here

$$
sum_ngeqslant1 dfraczeta(2n)n2^2n=colorbluelndfracpi2
$$

hardly, since I have
$$intsum_ngeq 1zeta(2n),x^n-1dx=intleft(dfrac12x^n+1-dfracpi2 dfraccot(pi x)x^nright)dx$$
and after integration set $x=dfrac14$, but it seems so hard.

Any suggestion, thanks in advanced!

integration zeta-functions

share|cite|improve this question

edited Jul 20 at 13:54

asked Jul 20 at 12:57

Nosrati

19.5k41544

share|cite|improve this question

share|cite|improve this question

edited Jul 20 at 13:54

edited Jul 20 at 13:54

edited Jul 20 at 13:54

asked Jul 20 at 12:57

Nosrati

19.5k41544

asked Jul 20 at 12:57

Nosrati

19.5k41544

asked Jul 20 at 12:57

Nosrati

19.5k41544

19.5k41544

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
5
down vote



accepted

Using your formula we have
$$ sum limits_n=1^infty fraczeta(2n)n 2^2n = int limits_0^1/2 sum limits_n=1^infty 2 zeta(2n) x^2n-1 , mathrmd x = int limits_0^1/2 frac1-pi x cot(pi x)x , mathrmd x , .$$
Now let $pi x = t$ and integrate:
$$ sum limits_n=1^infty fraczeta(2n)n 2^2n = lim_varepsilon searrow 0 int limits_varepsilon^pi/2 left[frac1t - cot(t)right] , mathrmd t = lim_varepsilon searrow 0 left[lnleft(fractsin(t)right)right]_varepsilon^pi /2 = ln left(fracpi2right) , .$$

Alternatively you can of course compute the series directly using Wallis' product:
beginalign
sum limits_n=1^infty fraczeta(2n)n 2^2n &= sum limits_n=1^infty frac1n 2^2n sum limits_k=1^infty frac1k^2n = sum limits_k=1^infty sum limits_n=1^infty frac1n (4k^2)^n = sum limits_k=1^infty - lnleft(1-frac14k^2right) \
&= sum limits_k=1^infty ln left(frac4k^24k^2 -1right) = ln left(prod limits_k=1^infty frac4k^24k^2 -1 right) = ln left(fracpi2right) , .
endalign

share|cite|improve this answer

edited Jul 20 at 13:29

answered Jul 20 at 13:22

ComplexYetTrivial

2,607624

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perfect (+1) $$
  – Szeto
  Jul 20 at 13:29
  

  
  
  
  

add a comment | 

up vote
4
down vote

Another (similar) approach, just for fun. From the integral representation of the Riemann Zeta function $$zetaleft(sright)=frac1Gammaleft(sright)int_0^inftyfracu^s-1e^u-1du,,mathrmReleft(sright)>1$$ we have $$S=sum_ngeq1fraczetaleft(2nright)n4^n=sum_ngeq1frac1n4^nleft(2n-1right)!int_0^inftyfracu^2n-1e^u-1du=int_0^inftyfrace^u/2+e^-u/2-2uleft(e^u-1right)du$$ where the exchange is justified by the dominated convergence theorem. Then, by the Frullani's theorem, we get $$S=sum_mgeq1left(int_0^inftyfrace^-uleft(m-1/2right)-e^-muudx+int_0^inftyfrace^-uleft(1/2+mright)-e^-muudxright)$$ $$=-sum_mgeq1logleft(1-frac14m^2right)$$ and so the claim by the Wallis product.

share|cite|improve this answer

answered Jul 20 at 14:24

Marco Cantarini

28.7k23574

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Frullani Rules! (+1)
  – Mark Viola
  Jul 20 at 14:51
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857613%2fwhats-the-series-of-sum-n-geqslant1-dfrac-zeta2nn22n%23new-answer', 'question_page');

);

Post as a guest

Name Email

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

Using your formula we have
$$ sum limits_n=1^infty fraczeta(2n)n 2^2n = int limits_0^1/2 sum limits_n=1^infty 2 zeta(2n) x^2n-1 , mathrmd x = int limits_0^1/2 frac1-pi x cot(pi x)x , mathrmd x , .$$
Now let $pi x = t$ and integrate:
$$ sum limits_n=1^infty fraczeta(2n)n 2^2n = lim_varepsilon searrow 0 int limits_varepsilon^pi/2 left[frac1t - cot(t)right] , mathrmd t = lim_varepsilon searrow 0 left[lnleft(fractsin(t)right)right]_varepsilon^pi /2 = ln left(fracpi2right) , .$$

Alternatively you can of course compute the series directly using Wallis' product:
beginalign
sum limits_n=1^infty fraczeta(2n)n 2^2n &= sum limits_n=1^infty frac1n 2^2n sum limits_k=1^infty frac1k^2n = sum limits_k=1^infty sum limits_n=1^infty frac1n (4k^2)^n = sum limits_k=1^infty - lnleft(1-frac14k^2right) \
&= sum limits_k=1^infty ln left(frac4k^24k^2 -1right) = ln left(prod limits_k=1^infty frac4k^24k^2 -1 right) = ln left(fracpi2right) , .
endalign

share|cite|improve this answer

edited Jul 20 at 13:29

answered Jul 20 at 13:22

ComplexYetTrivial

2,607624

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perfect (+1) $$
  – Szeto
  Jul 20 at 13:29
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

Using your formula we have
$$ sum limits_n=1^infty fraczeta(2n)n 2^2n = int limits_0^1/2 sum limits_n=1^infty 2 zeta(2n) x^2n-1 , mathrmd x = int limits_0^1/2 frac1-pi x cot(pi x)x , mathrmd x , .$$
Now let $pi x = t$ and integrate:
$$ sum limits_n=1^infty fraczeta(2n)n 2^2n = lim_varepsilon searrow 0 int limits_varepsilon^pi/2 left[frac1t - cot(t)right] , mathrmd t = lim_varepsilon searrow 0 left[lnleft(fractsin(t)right)right]_varepsilon^pi /2 = ln left(fracpi2right) , .$$

Alternatively you can of course compute the series directly using Wallis' product:
beginalign
sum limits_n=1^infty fraczeta(2n)n 2^2n &= sum limits_n=1^infty frac1n 2^2n sum limits_k=1^infty frac1k^2n = sum limits_k=1^infty sum limits_n=1^infty frac1n (4k^2)^n = sum limits_k=1^infty - lnleft(1-frac14k^2right) \
&= sum limits_k=1^infty ln left(frac4k^24k^2 -1right) = ln left(prod limits_k=1^infty frac4k^24k^2 -1 right) = ln left(fracpi2right) , .
endalign

share|cite|improve this answer

edited Jul 20 at 13:29

answered Jul 20 at 13:22

ComplexYetTrivial

2,607624

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perfect (+1) $$
  – Szeto
  Jul 20 at 13:29
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

Using your formula we have
$$ sum limits_n=1^infty fraczeta(2n)n 2^2n = int limits_0^1/2 sum limits_n=1^infty 2 zeta(2n) x^2n-1 , mathrmd x = int limits_0^1/2 frac1-pi x cot(pi x)x , mathrmd x , .$$
Now let $pi x = t$ and integrate:
$$ sum limits_n=1^infty fraczeta(2n)n 2^2n = lim_varepsilon searrow 0 int limits_varepsilon^pi/2 left[frac1t - cot(t)right] , mathrmd t = lim_varepsilon searrow 0 left[lnleft(fractsin(t)right)right]_varepsilon^pi /2 = ln left(fracpi2right) , .$$

Alternatively you can of course compute the series directly using Wallis' product:
beginalign
sum limits_n=1^infty fraczeta(2n)n 2^2n &= sum limits_n=1^infty frac1n 2^2n sum limits_k=1^infty frac1k^2n = sum limits_k=1^infty sum limits_n=1^infty frac1n (4k^2)^n = sum limits_k=1^infty - lnleft(1-frac14k^2right) \
&= sum limits_k=1^infty ln left(frac4k^24k^2 -1right) = ln left(prod limits_k=1^infty frac4k^24k^2 -1 right) = ln left(fracpi2right) , .
endalign

share|cite|improve this answer

edited Jul 20 at 13:29

answered Jul 20 at 13:22

ComplexYetTrivial

2,607624

Using your formula we have
$$ sum limits_n=1^infty fraczeta(2n)n 2^2n = int limits_0^1/2 sum limits_n=1^infty 2 zeta(2n) x^2n-1 , mathrmd x = int limits_0^1/2 frac1-pi x cot(pi x)x , mathrmd x , .$$
Now let $pi x = t$ and integrate:
$$ sum limits_n=1^infty fraczeta(2n)n 2^2n = lim_varepsilon searrow 0 int limits_varepsilon^pi/2 left[frac1t - cot(t)right] , mathrmd t = lim_varepsilon searrow 0 left[lnleft(fractsin(t)right)right]_varepsilon^pi /2 = ln left(fracpi2right) , .$$

Alternatively you can of course compute the series directly using Wallis' product:
beginalign
sum limits_n=1^infty fraczeta(2n)n 2^2n &= sum limits_n=1^infty frac1n 2^2n sum limits_k=1^infty frac1k^2n = sum limits_k=1^infty sum limits_n=1^infty frac1n (4k^2)^n = sum limits_k=1^infty - lnleft(1-frac14k^2right) \
&= sum limits_k=1^infty ln left(frac4k^24k^2 -1right) = ln left(prod limits_k=1^infty frac4k^24k^2 -1 right) = ln left(fracpi2right) , .
endalign

share|cite|improve this answer

edited Jul 20 at 13:29

answered Jul 20 at 13:22

ComplexYetTrivial

2,607624

share|cite|improve this answer

share|cite|improve this answer

edited Jul 20 at 13:29

edited Jul 20 at 13:29

edited Jul 20 at 13:29

answered Jul 20 at 13:22

ComplexYetTrivial

2,607624

answered Jul 20 at 13:22

ComplexYetTrivial

2,607624

answered Jul 20 at 13:22

ComplexYetTrivial

2,607624

2,607624

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perfect (+1) $$
  – Szeto
  Jul 20 at 13:29
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perfect (+1) $$
  – Szeto
  Jul 20 at 13:29
  

  
  
  
  

Perfect (+1) $$
– Szeto
Jul 20 at 13:29

Perfect (+1) $$
– Szeto
Jul 20 at 13:29

add a comment | 

up vote
4
down vote

Another (similar) approach, just for fun. From the integral representation of the Riemann Zeta function $$zetaleft(sright)=frac1Gammaleft(sright)int_0^inftyfracu^s-1e^u-1du,,mathrmReleft(sright)>1$$ we have $$S=sum_ngeq1fraczetaleft(2nright)n4^n=sum_ngeq1frac1n4^nleft(2n-1right)!int_0^inftyfracu^2n-1e^u-1du=int_0^inftyfrace^u/2+e^-u/2-2uleft(e^u-1right)du$$ where the exchange is justified by the dominated convergence theorem. Then, by the Frullani's theorem, we get $$S=sum_mgeq1left(int_0^inftyfrace^-uleft(m-1/2right)-e^-muudx+int_0^inftyfrace^-uleft(1/2+mright)-e^-muudxright)$$ $$=-sum_mgeq1logleft(1-frac14m^2right)$$ and so the claim by the Wallis product.

share|cite|improve this answer

answered Jul 20 at 14:24

Marco Cantarini

28.7k23574

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Frullani Rules! (+1)
  – Mark Viola
  Jul 20 at 14:51
  

  
  
  
  

add a comment | 

up vote
4
down vote

Another (similar) approach, just for fun. From the integral representation of the Riemann Zeta function $$zetaleft(sright)=frac1Gammaleft(sright)int_0^inftyfracu^s-1e^u-1du,,mathrmReleft(sright)>1$$ we have $$S=sum_ngeq1fraczetaleft(2nright)n4^n=sum_ngeq1frac1n4^nleft(2n-1right)!int_0^inftyfracu^2n-1e^u-1du=int_0^inftyfrace^u/2+e^-u/2-2uleft(e^u-1right)du$$ where the exchange is justified by the dominated convergence theorem. Then, by the Frullani's theorem, we get $$S=sum_mgeq1left(int_0^inftyfrace^-uleft(m-1/2right)-e^-muudx+int_0^inftyfrace^-uleft(1/2+mright)-e^-muudxright)$$ $$=-sum_mgeq1logleft(1-frac14m^2right)$$ and so the claim by the Wallis product.

share|cite|improve this answer

answered Jul 20 at 14:24

Marco Cantarini

28.7k23574

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Frullani Rules! (+1)
  – Mark Viola
  Jul 20 at 14:51
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

Another (similar) approach, just for fun. From the integral representation of the Riemann Zeta function $$zetaleft(sright)=frac1Gammaleft(sright)int_0^inftyfracu^s-1e^u-1du,,mathrmReleft(sright)>1$$ we have $$S=sum_ngeq1fraczetaleft(2nright)n4^n=sum_ngeq1frac1n4^nleft(2n-1right)!int_0^inftyfracu^2n-1e^u-1du=int_0^inftyfrace^u/2+e^-u/2-2uleft(e^u-1right)du$$ where the exchange is justified by the dominated convergence theorem. Then, by the Frullani's theorem, we get $$S=sum_mgeq1left(int_0^inftyfrace^-uleft(m-1/2right)-e^-muudx+int_0^inftyfrace^-uleft(1/2+mright)-e^-muudxright)$$ $$=-sum_mgeq1logleft(1-frac14m^2right)$$ and so the claim by the Wallis product.

share|cite|improve this answer

answered Jul 20 at 14:24

Marco Cantarini

28.7k23574

Another (similar) approach, just for fun. From the integral representation of the Riemann Zeta function $$zetaleft(sright)=frac1Gammaleft(sright)int_0^inftyfracu^s-1e^u-1du,,mathrmReleft(sright)>1$$ we have $$S=sum_ngeq1fraczetaleft(2nright)n4^n=sum_ngeq1frac1n4^nleft(2n-1right)!int_0^inftyfracu^2n-1e^u-1du=int_0^inftyfrace^u/2+e^-u/2-2uleft(e^u-1right)du$$ where the exchange is justified by the dominated convergence theorem. Then, by the Frullani's theorem, we get $$S=sum_mgeq1left(int_0^inftyfrace^-uleft(m-1/2right)-e^-muudx+int_0^inftyfrace^-uleft(1/2+mright)-e^-muudxright)$$ $$=-sum_mgeq1logleft(1-frac14m^2right)$$ and so the claim by the Wallis product.

share|cite|improve this answer

answered Jul 20 at 14:24

Marco Cantarini

28.7k23574

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 14:24

Marco Cantarini

28.7k23574

answered Jul 20 at 14:24

Marco Cantarini

28.7k23574

answered Jul 20 at 14:24

Marco Cantarini

28.7k23574

28.7k23574

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Frullani Rules! (+1)
  – Mark Viola
  Jul 20 at 14:51
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Frullani Rules! (+1)
  – Mark Viola
  Jul 20 at 14:51
  

  
  
  
  

2

2

Frullani Rules! (+1)
– Mark Viola
Jul 20 at 14:51

Frullani Rules! (+1)
– Mark Viola
Jul 20 at 14:51

add a comment | 

 

draft saved

draft discarded

 

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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857613%2fwhats-the-series-of-sum-n-geqslant1-dfrac-zeta2nn22n%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

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

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

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email