Mixing
How to prove that $frac{zeta(2) }{2}+frac{zeta (4)}{2^3}+frac{zeta (6)}{2^5}+frac{zeta (8)}{2^7}+cdots=1$?
How can one prove this identity?
$$frac{zeta(2) }{2}+frac{zeta (4)}{2^3}+frac{zeta (6)}{2^5}+frac{zeta (8)}{2^7}+cdots=1$$
There is a formula for $zeta$ values at even integers, but it involves Bernoulli numbers; simply plugging it in does not appear to be an efficient approach.
real-analysis sequences-and-series complex-analysis zeta-functions
asked Dec 24 '14 at 18:24
E.H.E
15.7k11966
add a comment |
How can one prove this identity?
$$frac{zeta(2) }{2}+frac{zeta (4)}{2^3}+frac{zeta (6)}{2^5}+frac{zeta (8)}{2^7}+cdots=1$$
There is a formula for $zeta$ values at even integers, but it involves Bernoulli numbers; simply plugging it in does not appear to be an efficient approach.
real-analysis sequences-and-series complex-analysis zeta-functions
asked Dec 24 '14 at 18:24
E.H.E
15.7k11966
12
I think this is a normal question. I don't know why "on hold"?
– E.H.E
Dec 24 '14 at 19:28
2
There is a recursive formula for $zeta(2n)$ derived in this answer. However, in that answer it is shown that $$sum_{k=1}^inftyzeta(2k)x^{2k}=frac12(1-pi xcot(pi x))$$ With $x=frac12$, this immediately gives $$sum_{k=1}^inftyzeta(2k)left(frac12right)^{2k-1} =1-fracpi2cotleft(fracpi2right)=1$$ This is what is used in Random Variable's answer.
– robjohn♦
Dec 31 '14 at 18:13
add a comment |
How can one prove this identity?
$$frac{zeta(2) }{2}+frac{zeta (4)}{2^3}+frac{zeta (6)}{2^5}+frac{zeta (8)}{2^7}+cdots=1$$
There is a formula for $zeta$ values at even integers, but it involves Bernoulli numbers; simply plugging it in does not appear to be an efficient approach.
real-analysis sequences-and-series complex-analysis zeta-functions
asked Dec 24 '14 at 18:24
E.H.E
15.7k11966
How can one prove this identity?
$$frac{zeta(2) }{2}+frac{zeta (4)}{2^3}+frac{zeta (6)}{2^5}+frac{zeta (8)}{2^7}+cdots=1$$
There is a formula for $zeta$ values at even integers, but it involves Bernoulli numbers; simply plugging it in does not appear to be an efficient approach.
real-analysis sequences-and-series complex-analysis zeta-functions
real-analysis sequences-and-series complex-analysis zeta-functions
asked Dec 24 '14 at 18:24
E.H.E
15.7k11966
asked Dec 24 '14 at 18:24
E.H.E
15.7k11966
edited Dec 25 '14 at 1:05
user147263
asked Dec 24 '14 at 18:24
E.H.E
15.7k11966
asked Dec 24 '14 at 18:24
E.H.E
15.7k11966
asked Dec 24 '14 at 18:24
E.H.E
15.7k11966
15.7k11966
12
I think this is a normal question. I don't know why "on hold"?
– E.H.E
Dec 24 '14 at 19:28
2
There is a recursive formula for $zeta(2n)$ derived in this answer. However, in that answer it is shown that $$sum_{k=1}^inftyzeta(2k)x^{2k}=frac12(1-pi xcot(pi x))$$ With $x=frac12$, this immediately gives $$sum_{k=1}^inftyzeta(2k)left(frac12right)^{2k-1} =1-fracpi2cotleft(fracpi2right)=1$$ This is what is used in Random Variable's answer.
– robjohn♦
Dec 31 '14 at 18:13
add a comment |
12
I think this is a normal question. I don't know why "on hold"?
– E.H.E
Dec 24 '14 at 19:28
2
There is a recursive formula for $zeta(2n)$ derived in this answer. However, in that answer it is shown that $$sum_{k=1}^inftyzeta(2k)x^{2k}=frac12(1-pi xcot(pi x))$$ With $x=frac12$, this immediately gives $$sum_{k=1}^inftyzeta(2k)left(frac12right)^{2k-1} =1-fracpi2cotleft(fracpi2right)=1$$ This is what is used in Random Variable's answer.
– robjohn♦
Dec 31 '14 at 18:13
12
12
I think this is a normal question. I don't know why "on hold"?
– E.H.E
Dec 24 '14 at 19:28
I think this is a normal question. I don't know why "on hold"?
– E.H.E
Dec 24 '14 at 19:28
2
2
There is a recursive formula for $zeta(2n)$ derived in this answer. However, in that answer it is shown that $$sum_{k=1}^inftyzeta(2k)x^{2k}=frac12(1-pi xcot(pi x))$$ With $x=frac12$, this immediately gives $$sum_{k=1}^inftyzeta(2k)left(frac12right)^{2k-1} =1-fracpi2cotleft(fracpi2right)=1$$ This is what is used in Random Variable's answer.
– robjohn♦
Dec 31 '14 at 18:13
There is a recursive formula for $zeta(2n)$ derived in this answer. However, in that answer it is shown that $$sum_{k=1}^inftyzeta(2k)x^{2k}=frac12(1-pi xcot(pi x))$$ With $x=frac12$, this immediately gives $$sum_{k=1}^inftyzeta(2k)left(frac12right)^{2k-1} =1-fracpi2cotleft(fracpi2right)=1$$ This is what is used in Random Variable's answer.
– robjohn♦
Dec 31 '14 at 18:13
add a comment |
4 Answers
4
active
oldest
votes
Since
$$zeta(2n) = frac{1}{(2n-1)!}int_{0}^{+infty}frac{x^{2n-1}}{e^x-1},dx $$
we have:
$$sum_{n=1}^{+infty}frac{zeta(2n)}{2^{2n-1}} = int_{0}^{+infty}frac{sinh(x/2)}{e^x-1},dx =frac{1}{2}int_{0}^{+infty}e^{-x/2},dx = color{red}{1}.$$
answered Dec 24 '14 at 18:29
Jack D'Aurizio
286k33279656
2
@mathworker21: I can mention that the dominated (or just monotone) convergence theorem clearly applies here.
– Jack D'Aurizio
Jul 14 '17 at 18:41
add a comment |
$$
begin{align}
sum_{n=1}^inftyfrac{zeta(2n)}{2^{2n-1}}
&=sum_{n=1}^inftysum_{k=1}^inftyfrac2{k^{2n}2^{2n}}tag{1}\
&=2sum_{k=1}^inftysum_{n=1}^inftyfrac1{(4k^2)^n}tag{2}\
&=2sum_{k=1}^inftyfrac1{4k^2-1}tag{3}\
&=sum_{k=1}^inftyleft(frac1{2k-1}-frac1{2k+1}right)tag{4}\[6pt]
&=1tag{5}
end{align}
$$
Explanation:
$(1)$: expand $zeta(2n)=sumlimits_{k=1}^inftyfrac1{k^{2n}}$
$(2)$: change the order of summation
$(3)$: sum of a geometric series
$(4)$: partial fractions
$(5)$: telescoping sum
answered Dec 24 '14 at 20:44
robjohn♦
264k27303623
@robjohn, in $(3)$ the numerator should have a $4k^2$. $$sum_{n=1}^{infty} frac{1}{(4k^2)^n} = frac{1}{1- frac{1}{4k^2}} = frac{4k^2}{4k^2 - 1}$$
– Amad27
Mar 20 '15 at 16:45
2
@Amad27: Notice that the sum starts from $n=1$. This means that the sum is $dfrac{frac1{4k^2}}{1-frac1{4k^2}}$.
– robjohn♦
Mar 20 '15 at 17:10
Ah! I couldn't catch the index! Sorry robjohn!
– Amad27
Mar 20 '15 at 18:17
1
This one appears to be simplest possible answer. +1 It is as if you see the result in question at the level of $1 + r + r^{2} + cdots = 1/(1 - r)$.
– Paramanand Singh
Jun 20 '16 at 10:29
add a comment |
The Laurent expansion of $cot (z)$ at the origin in terms of the Riemann zeta function is $$ cot (z) = frac{1}{z} - 2 sum_{k=1}^{infty}zeta(2k) frac{z^{2k-1}}{pi^{2k}} , 0 < |z| < pi. $$
Letting $ displaystyle z= frac{pi}{2}$, $$cot left(frac{pi}{2} right) = frac{2}{pi} - frac{2}{pi} sum_{k=1}^{infty} frac{zeta(2k)}{2^{2k-1}}.$$
But $cot left(frac{pi}{2} right)=0$.
Therefore,
$$ sum_{k=1}^{infty} frac{zeta(2k)}{2^{2k-1}} = 1.$$
answered Dec 24 '14 at 22:27
Random Variable
25.4k170136
+1. Nice. I was doing something like that with identities $color{black}{bf 6.3.14}$ and $color{black}{bf 6.3.15}$ but I left because it was cumbersome.
– Felix Marin
Dec 24 '14 at 22:54
@FelixMarin Thanks. This actually wasn't my first approach.
– Random Variable
Dec 24 '14 at 23:20
(+1) Beautiful approach! This relation is also proven in this answer.
– robjohn♦
Dec 25 '14 at 0:06
add a comment |
$newcommand{angles}[1]{leftlangle{#1}rightrangle}
newcommand{braces}[1]{leftlbrace{#1}rightrbrace}
newcommand{bracks}[1]{leftlbrack{#1}rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Leftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left({#1}right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert{#1}rightvert}$
begin{align}
&bbox[10px,#ffd]{sum_{n = 1}^{infty}{zetapars{2n} over 2^{2n - 1}}} =
sum_{n = 2}^{infty}{zetapars{n} over 2^{n - 1}} - sum_{n = 1}^{infty}{zetapars{2n + 1} over 2^{2n}}
\[3mm] = &
-sum_{n = 2}^{infty}pars{-1}^{n}zetapars{n}pars{-,half}^{n - 1} -
sum_{n = 1}^{infty}bracks{zetapars{2n + 1} - 1}pars{half}^{2n} -
underbrace{sum_{n = 1}^{infty}pars{half}^{2n}}_{ds{1 over 3}}
\ = &
-bracks{Psipars{1 + z} + gamma}_{ z = -1/2}
\[3mm] & - bracks{%
{1 over 2z} - half,picotpars{pi z} - {1 over 1 - z^{2}} + 1 - gamma - Psipars{1 + z}}_{ z = 1/2} - {1 over 3}
\[8mm] = &
-Psipars{half} - {2 over 3} +
underbrace{Psipars{3 over 2}}_{ds{Psipars{1/2} + 1/pars{1/2}}} -
{1 over 3} = color{#f00}{1}
end{align}
$Psi$ and $gamma$ are the Digamma function and the Euler-Mascheroni constant, respectively. We used the Digamma Recurrence Formula $ds{Psipars{z + 1} = Psipars{z} + 1/z}$ and the identities $mathbf{6.3.14}$ and $mathbf{6.3.15}$ in this link.
answered Jun 19 '16 at 19:42
Felix Marin
67.1k7107141
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',
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1080000%2fhow-to-prove-that-frac-zeta2-2-frac-zeta-423-frac-zeta-62%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Since
$$zeta(2n) = frac{1}{(2n-1)!}int_{0}^{+infty}frac{x^{2n-1}}{e^x-1},dx $$
we have:
$$sum_{n=1}^{+infty}frac{zeta(2n)}{2^{2n-1}} = int_{0}^{+infty}frac{sinh(x/2)}{e^x-1},dx =frac{1}{2}int_{0}^{+infty}e^{-x/2},dx = color{red}{1}.$$
answered Dec 24 '14 at 18:29
Jack D'Aurizio
286k33279656
2
@mathworker21: I can mention that the dominated (or just monotone) convergence theorem clearly applies here.
– Jack D'Aurizio
Jul 14 '17 at 18:41
add a comment |
Since
$$zeta(2n) = frac{1}{(2n-1)!}int_{0}^{+infty}frac{x^{2n-1}}{e^x-1},dx $$
we have:
$$sum_{n=1}^{+infty}frac{zeta(2n)}{2^{2n-1}} = int_{0}^{+infty}frac{sinh(x/2)}{e^x-1},dx =frac{1}{2}int_{0}^{+infty}e^{-x/2},dx = color{red}{1}.$$
answered Dec 24 '14 at 18:29
Jack D'Aurizio
286k33279656
2
@mathworker21: I can mention that the dominated (or just monotone) convergence theorem clearly applies here.
– Jack D'Aurizio
Jul 14 '17 at 18:41
add a comment |
Since
$$zeta(2n) = frac{1}{(2n-1)!}int_{0}^{+infty}frac{x^{2n-1}}{e^x-1},dx $$
we have:
$$sum_{n=1}^{+infty}frac{zeta(2n)}{2^{2n-1}} = int_{0}^{+infty}frac{sinh(x/2)}{e^x-1},dx =frac{1}{2}int_{0}^{+infty}e^{-x/2},dx = color{red}{1}.$$
answered Dec 24 '14 at 18:29
Jack D'Aurizio
286k33279656
Since
$$zeta(2n) = frac{1}{(2n-1)!}int_{0}^{+infty}frac{x^{2n-1}}{e^x-1},dx $$
we have:
$$sum_{n=1}^{+infty}frac{zeta(2n)}{2^{2n-1}} = int_{0}^{+infty}frac{sinh(x/2)}{e^x-1},dx =frac{1}{2}int_{0}^{+infty}e^{-x/2},dx = color{red}{1}.$$
answered Dec 24 '14 at 18:29
Jack D'Aurizio
286k33279656
answered Dec 24 '14 at 18:29
Jack D'Aurizio
286k33279656
answered Dec 24 '14 at 18:29
Jack D'Aurizio
286k33279656
answered Dec 24 '14 at 18:29
Jack D'Aurizio
286k33279656
286k33279656
2
@mathworker21: I can mention that the dominated (or just monotone) convergence theorem clearly applies here.
– Jack D'Aurizio
Jul 14 '17 at 18:41
add a comment |
2
@mathworker21: I can mention that the dominated (or just monotone) convergence theorem clearly applies here.
– Jack D'Aurizio
Jul 14 '17 at 18:41
2
2
@mathworker21: I can mention that the dominated (or just monotone) convergence theorem clearly applies here.
– Jack D'Aurizio
Jul 14 '17 at 18:41
@mathworker21: I can mention that the dominated (or just monotone) convergence theorem clearly applies here.
– Jack D'Aurizio
Jul 14 '17 at 18:41
add a comment |
$$
begin{align}
sum_{n=1}^inftyfrac{zeta(2n)}{2^{2n-1}}
&=sum_{n=1}^inftysum_{k=1}^inftyfrac2{k^{2n}2^{2n}}tag{1}\
&=2sum_{k=1}^inftysum_{n=1}^inftyfrac1{(4k^2)^n}tag{2}\
&=2sum_{k=1}^inftyfrac1{4k^2-1}tag{3}\
&=sum_{k=1}^inftyleft(frac1{2k-1}-frac1{2k+1}right)tag{4}\[6pt]
&=1tag{5}
end{align}
$$
Explanation:
$(1)$: expand $zeta(2n)=sumlimits_{k=1}^inftyfrac1{k^{2n}}$
$(2)$: change the order of summation
$(3)$: sum of a geometric series
$(4)$: partial fractions
$(5)$: telescoping sum
answered Dec 24 '14 at 20:44
robjohn♦
264k27303623
@robjohn, in $(3)$ the numerator should have a $4k^2$. $$sum_{n=1}^{infty} frac{1}{(4k^2)^n} = frac{1}{1- frac{1}{4k^2}} = frac{4k^2}{4k^2 - 1}$$
– Amad27
Mar 20 '15 at 16:45
2
@Amad27: Notice that the sum starts from $n=1$. This means that the sum is $dfrac{frac1{4k^2}}{1-frac1{4k^2}}$.
– robjohn♦
Mar 20 '15 at 17:10
Ah! I couldn't catch the index! Sorry robjohn!
– Amad27
Mar 20 '15 at 18:17
1
This one appears to be simplest possible answer. +1 It is as if you see the result in question at the level of $1 + r + r^{2} + cdots = 1/(1 - r)$.
– Paramanand Singh
Jun 20 '16 at 10:29
add a comment |
$$
begin{align}
sum_{n=1}^inftyfrac{zeta(2n)}{2^{2n-1}}
&=sum_{n=1}^inftysum_{k=1}^inftyfrac2{k^{2n}2^{2n}}tag{1}\
&=2sum_{k=1}^inftysum_{n=1}^inftyfrac1{(4k^2)^n}tag{2}\
&=2sum_{k=1}^inftyfrac1{4k^2-1}tag{3}\
&=sum_{k=1}^inftyleft(frac1{2k-1}-frac1{2k+1}right)tag{4}\[6pt]
&=1tag{5}
end{align}
$$
Explanation:
$(1)$: expand $zeta(2n)=sumlimits_{k=1}^inftyfrac1{k^{2n}}$
$(2)$: change the order of summation
$(3)$: sum of a geometric series
$(4)$: partial fractions
$(5)$: telescoping sum
answered Dec 24 '14 at 20:44
robjohn♦
264k27303623
@robjohn, in $(3)$ the numerator should have a $4k^2$. $$sum_{n=1}^{infty} frac{1}{(4k^2)^n} = frac{1}{1- frac{1}{4k^2}} = frac{4k^2}{4k^2 - 1}$$
– Amad27
Mar 20 '15 at 16:45
2
@Amad27: Notice that the sum starts from $n=1$. This means that the sum is $dfrac{frac1{4k^2}}{1-frac1{4k^2}}$.
– robjohn♦
Mar 20 '15 at 17:10
Ah! I couldn't catch the index! Sorry robjohn!
– Amad27
Mar 20 '15 at 18:17
1
This one appears to be simplest possible answer. +1 It is as if you see the result in question at the level of $1 + r + r^{2} + cdots = 1/(1 - r)$.
– Paramanand Singh
Jun 20 '16 at 10:29
add a comment |
$$
begin{align}
sum_{n=1}^inftyfrac{zeta(2n)}{2^{2n-1}}
&=sum_{n=1}^inftysum_{k=1}^inftyfrac2{k^{2n}2^{2n}}tag{1}\
&=2sum_{k=1}^inftysum_{n=1}^inftyfrac1{(4k^2)^n}tag{2}\
&=2sum_{k=1}^inftyfrac1{4k^2-1}tag{3}\
&=sum_{k=1}^inftyleft(frac1{2k-1}-frac1{2k+1}right)tag{4}\[6pt]
&=1tag{5}
end{align}
$$
Explanation:
$(1)$: expand $zeta(2n)=sumlimits_{k=1}^inftyfrac1{k^{2n}}$
$(2)$: change the order of summation
$(3)$: sum of a geometric series
$(4)$: partial fractions
$(5)$: telescoping sum
answered Dec 24 '14 at 20:44
robjohn♦
264k27303623
$$
begin{align}
sum_{n=1}^inftyfrac{zeta(2n)}{2^{2n-1}}
&=sum_{n=1}^inftysum_{k=1}^inftyfrac2{k^{2n}2^{2n}}tag{1}\
&=2sum_{k=1}^inftysum_{n=1}^inftyfrac1{(4k^2)^n}tag{2}\
&=2sum_{k=1}^inftyfrac1{4k^2-1}tag{3}\
&=sum_{k=1}^inftyleft(frac1{2k-1}-frac1{2k+1}right)tag{4}\[6pt]
&=1tag{5}
end{align}
$$
Explanation:
$(1)$: expand $zeta(2n)=sumlimits_{k=1}^inftyfrac1{k^{2n}}$
$(2)$: change the order of summation
$(3)$: sum of a geometric series
$(4)$: partial fractions
$(5)$: telescoping sum
answered Dec 24 '14 at 20:44
robjohn♦
264k27303623
edited Dec 24 '14 at 20:49
answered Dec 24 '14 at 20:44
robjohn♦
264k27303623
answered Dec 24 '14 at 20:44
robjohn♦
264k27303623
answered Dec 24 '14 at 20:44
robjohn♦
264k27303623
264k27303623
@robjohn, in $(3)$ the numerator should have a $4k^2$. $$sum_{n=1}^{infty} frac{1}{(4k^2)^n} = frac{1}{1- frac{1}{4k^2}} = frac{4k^2}{4k^2 - 1}$$
– Amad27
Mar 20 '15 at 16:45
2
@Amad27: Notice that the sum starts from $n=1$. This means that the sum is $dfrac{frac1{4k^2}}{1-frac1{4k^2}}$.
– robjohn♦
Mar 20 '15 at 17:10
Ah! I couldn't catch the index! Sorry robjohn!
– Amad27
Mar 20 '15 at 18:17
1
This one appears to be simplest possible answer. +1 It is as if you see the result in question at the level of $1 + r + r^{2} + cdots = 1/(1 - r)$.
– Paramanand Singh
Jun 20 '16 at 10:29
add a comment |
@robjohn, in $(3)$ the numerator should have a $4k^2$. $$sum_{n=1}^{infty} frac{1}{(4k^2)^n} = frac{1}{1- frac{1}{4k^2}} = frac{4k^2}{4k^2 - 1}$$
– Amad27
Mar 20 '15 at 16:45
2
@Amad27: Notice that the sum starts from $n=1$. This means that the sum is $dfrac{frac1{4k^2}}{1-frac1{4k^2}}$.
– robjohn♦
Mar 20 '15 at 17:10
Ah! I couldn't catch the index! Sorry robjohn!
– Amad27
Mar 20 '15 at 18:17
1
This one appears to be simplest possible answer. +1 It is as if you see the result in question at the level of $1 + r + r^{2} + cdots = 1/(1 - r)$.
– Paramanand Singh
Jun 20 '16 at 10:29
@robjohn, in $(3)$ the numerator should have a $4k^2$. $$sum_{n=1}^{infty} frac{1}{(4k^2)^n} = frac{1}{1- frac{1}{4k^2}} = frac{4k^2}{4k^2 - 1}$$
– Amad27
Mar 20 '15 at 16:45
@robjohn, in $(3)$ the numerator should have a $4k^2$. $$sum_{n=1}^{infty} frac{1}{(4k^2)^n} = frac{1}{1- frac{1}{4k^2}} = frac{4k^2}{4k^2 - 1}$$
– Amad27
Mar 20 '15 at 16:45
2
2
@Amad27: Notice that the sum starts from $n=1$. This means that the sum is $dfrac{frac1{4k^2}}{1-frac1{4k^2}}$.
– robjohn♦
Mar 20 '15 at 17:10
@Amad27: Notice that the sum starts from $n=1$. This means that the sum is $dfrac{frac1{4k^2}}{1-frac1{4k^2}}$.
– robjohn♦
Mar 20 '15 at 17:10
Ah! I couldn't catch the index! Sorry robjohn!
– Amad27
Mar 20 '15 at 18:17
Ah! I couldn't catch the index! Sorry robjohn!
– Amad27
Mar 20 '15 at 18:17
1
1
This one appears to be simplest possible answer. +1 It is as if you see the result in question at the level of $1 + r + r^{2} + cdots = 1/(1 - r)$.
– Paramanand Singh
Jun 20 '16 at 10:29
This one appears to be simplest possible answer. +1 It is as if you see the result in question at the level of $1 + r + r^{2} + cdots = 1/(1 - r)$.
– Paramanand Singh
Jun 20 '16 at 10:29
add a comment |
The Laurent expansion of $cot (z)$ at the origin in terms of the Riemann zeta function is $$ cot (z) = frac{1}{z} - 2 sum_{k=1}^{infty}zeta(2k) frac{z^{2k-1}}{pi^{2k}} , 0 < |z| < pi. $$
Letting $ displaystyle z= frac{pi}{2}$, $$cot left(frac{pi}{2} right) = frac{2}{pi} - frac{2}{pi} sum_{k=1}^{infty} frac{zeta(2k)}{2^{2k-1}}.$$
But $cot left(frac{pi}{2} right)=0$.
Therefore,
$$ sum_{k=1}^{infty} frac{zeta(2k)}{2^{2k-1}} = 1.$$
answered Dec 24 '14 at 22:27
Random Variable
25.4k170136
+1. Nice. I was doing something like that with identities $color{black}{bf 6.3.14}$ and $color{black}{bf 6.3.15}$ but I left because it was cumbersome.
– Felix Marin
Dec 24 '14 at 22:54
@FelixMarin Thanks. This actually wasn't my first approach.
– Random Variable
Dec 24 '14 at 23:20
(+1) Beautiful approach! This relation is also proven in this answer.
– robjohn♦
Dec 25 '14 at 0:06
add a comment |
The Laurent expansion of $cot (z)$ at the origin in terms of the Riemann zeta function is $$ cot (z) = frac{1}{z} - 2 sum_{k=1}^{infty}zeta(2k) frac{z^{2k-1}}{pi^{2k}} , 0 < |z| < pi. $$
Letting $ displaystyle z= frac{pi}{2}$, $$cot left(frac{pi}{2} right) = frac{2}{pi} - frac{2}{pi} sum_{k=1}^{infty} frac{zeta(2k)}{2^{2k-1}}.$$
But $cot left(frac{pi}{2} right)=0$.
Therefore,
$$ sum_{k=1}^{infty} frac{zeta(2k)}{2^{2k-1}} = 1.$$
answered Dec 24 '14 at 22:27
Random Variable
25.4k170136
+1. Nice. I was doing something like that with identities $color{black}{bf 6.3.14}$ and $color{black}{bf 6.3.15}$ but I left because it was cumbersome.
– Felix Marin
Dec 24 '14 at 22:54
@FelixMarin Thanks. This actually wasn't my first approach.
– Random Variable
Dec 24 '14 at 23:20
(+1) Beautiful approach! This relation is also proven in this answer.
– robjohn♦
Dec 25 '14 at 0:06
add a comment |
The Laurent expansion of $cot (z)$ at the origin in terms of the Riemann zeta function is $$ cot (z) = frac{1}{z} - 2 sum_{k=1}^{infty}zeta(2k) frac{z^{2k-1}}{pi^{2k}} , 0 < |z| < pi. $$
Letting $ displaystyle z= frac{pi}{2}$, $$cot left(frac{pi}{2} right) = frac{2}{pi} - frac{2}{pi} sum_{k=1}^{infty} frac{zeta(2k)}{2^{2k-1}}.$$
But $cot left(frac{pi}{2} right)=0$.
Therefore,
$$ sum_{k=1}^{infty} frac{zeta(2k)}{2^{2k-1}} = 1.$$
answered Dec 24 '14 at 22:27
Random Variable
25.4k170136
The Laurent expansion of $cot (z)$ at the origin in terms of the Riemann zeta function is $$ cot (z) = frac{1}{z} - 2 sum_{k=1}^{infty}zeta(2k) frac{z^{2k-1}}{pi^{2k}} , 0 < |z| < pi. $$
Letting $ displaystyle z= frac{pi}{2}$, $$cot left(frac{pi}{2} right) = frac{2}{pi} - frac{2}{pi} sum_{k=1}^{infty} frac{zeta(2k)}{2^{2k-1}}.$$
But $cot left(frac{pi}{2} right)=0$.
Therefore,
$$ sum_{k=1}^{infty} frac{zeta(2k)}{2^{2k-1}} = 1.$$
answered Dec 24 '14 at 22:27
Random Variable
25.4k170136
edited Dec 24 '14 at 22:33
answered Dec 24 '14 at 22:27
Random Variable
25.4k170136
answered Dec 24 '14 at 22:27
Random Variable
25.4k170136
answered Dec 24 '14 at 22:27
Random Variable
25.4k170136
25.4k170136
+1. Nice. I was doing something like that with identities $color{black}{bf 6.3.14}$ and $color{black}{bf 6.3.15}$ but I left because it was cumbersome.
– Felix Marin
Dec 24 '14 at 22:54
@FelixMarin Thanks. This actually wasn't my first approach.
– Random Variable
Dec 24 '14 at 23:20
(+1) Beautiful approach! This relation is also proven in this answer.
– robjohn♦
Dec 25 '14 at 0:06
add a comment |
+1. Nice. I was doing something like that with identities $color{black}{bf 6.3.14}$ and $color{black}{bf 6.3.15}$ but I left because it was cumbersome.
– Felix Marin
Dec 24 '14 at 22:54
@FelixMarin Thanks. This actually wasn't my first approach.
– Random Variable
Dec 24 '14 at 23:20
(+1) Beautiful approach! This relation is also proven in this answer.
– robjohn♦
Dec 25 '14 at 0:06
+1. Nice. I was doing something like that with identities $color{black}{bf 6.3.14}$ and $color{black}{bf 6.3.15}$ but I left because it was cumbersome.
– Felix Marin
Dec 24 '14 at 22:54
+1. Nice. I was doing something like that with identities $color{black}{bf 6.3.14}$ and $color{black}{bf 6.3.15}$ but I left because it was cumbersome.
– Felix Marin
Dec 24 '14 at 22:54
@FelixMarin Thanks. This actually wasn't my first approach.
– Random Variable
Dec 24 '14 at 23:20
@FelixMarin Thanks. This actually wasn't my first approach.
– Random Variable
Dec 24 '14 at 23:20
(+1) Beautiful approach! This relation is also proven in this answer.
– robjohn♦
Dec 25 '14 at 0:06
(+1) Beautiful approach! This relation is also proven in this answer.
– robjohn♦
Dec 25 '14 at 0:06
add a comment |
$newcommand{angles}[1]{leftlangle{#1}rightrangle}
newcommand{braces}[1]{leftlbrace{#1}rightrbrace}
newcommand{bracks}[1]{leftlbrack{#1}rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Leftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left({#1}right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert{#1}rightvert}$
begin{align}
&bbox[10px,#ffd]{sum_{n = 1}^{infty}{zetapars{2n} over 2^{2n - 1}}} =
sum_{n = 2}^{infty}{zetapars{n} over 2^{n - 1}} - sum_{n = 1}^{infty}{zetapars{2n + 1} over 2^{2n}}
\[3mm] = &
-sum_{n = 2}^{infty}pars{-1}^{n}zetapars{n}pars{-,half}^{n - 1} -
sum_{n = 1}^{infty}bracks{zetapars{2n + 1} - 1}pars{half}^{2n} -
underbrace{sum_{n = 1}^{infty}pars{half}^{2n}}_{ds{1 over 3}}
\ = &
-bracks{Psipars{1 + z} + gamma}_{ z = -1/2}
\[3mm] & - bracks{%
{1 over 2z} - half,picotpars{pi z} - {1 over 1 - z^{2}} + 1 - gamma - Psipars{1 + z}}_{ z = 1/2} - {1 over 3}
\[8mm] = &
-Psipars{half} - {2 over 3} +
underbrace{Psipars{3 over 2}}_{ds{Psipars{1/2} + 1/pars{1/2}}} -
{1 over 3} = color{#f00}{1}
end{align}
$Psi$ and $gamma$ are the Digamma function and the Euler-Mascheroni constant, respectively. We used the Digamma Recurrence Formula $ds{Psipars{z + 1} = Psipars{z} + 1/z}$ and the identities $mathbf{6.3.14}$ and $mathbf{6.3.15}$ in this link.
answered Jun 19 '16 at 19:42
Felix Marin
67.1k7107141
add a comment |
$newcommand{angles}[1]{leftlangle{#1}rightrangle}
newcommand{braces}[1]{leftlbrace{#1}rightrbrace}
newcommand{bracks}[1]{leftlbrack{#1}rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Leftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left({#1}right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert{#1}rightvert}$
begin{align}
&bbox[10px,#ffd]{sum_{n = 1}^{infty}{zetapars{2n} over 2^{2n - 1}}} =
sum_{n = 2}^{infty}{zetapars{n} over 2^{n - 1}} - sum_{n = 1}^{infty}{zetapars{2n + 1} over 2^{2n}}
\[3mm] = &
-sum_{n = 2}^{infty}pars{-1}^{n}zetapars{n}pars{-,half}^{n - 1} -
sum_{n = 1}^{infty}bracks{zetapars{2n + 1} - 1}pars{half}^{2n} -
underbrace{sum_{n = 1}^{infty}pars{half}^{2n}}_{ds{1 over 3}}
\ = &
-bracks{Psipars{1 + z} + gamma}_{ z = -1/2}
\[3mm] & - bracks{%
{1 over 2z} - half,picotpars{pi z} - {1 over 1 - z^{2}} + 1 - gamma - Psipars{1 + z}}_{ z = 1/2} - {1 over 3}
\[8mm] = &
-Psipars{half} - {2 over 3} +
underbrace{Psipars{3 over 2}}_{ds{Psipars{1/2} + 1/pars{1/2}}} -
{1 over 3} = color{#f00}{1}
end{align}
$Psi$ and $gamma$ are the Digamma function and the Euler-Mascheroni constant, respectively. We used the Digamma Recurrence Formula $ds{Psipars{z + 1} = Psipars{z} + 1/z}$ and the identities $mathbf{6.3.14}$ and $mathbf{6.3.15}$ in this link.
answered Jun 19 '16 at 19:42
Felix Marin
67.1k7107141
add a comment |
$newcommand{angles}[1]{leftlangle{#1}rightrangle}
newcommand{braces}[1]{leftlbrace{#1}rightrbrace}
newcommand{bracks}[1]{leftlbrack{#1}rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Leftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left({#1}right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert{#1}rightvert}$
begin{align}
&bbox[10px,#ffd]{sum_{n = 1}^{infty}{zetapars{2n} over 2^{2n - 1}}} =
sum_{n = 2}^{infty}{zetapars{n} over 2^{n - 1}} - sum_{n = 1}^{infty}{zetapars{2n + 1} over 2^{2n}}
\[3mm] = &
-sum_{n = 2}^{infty}pars{-1}^{n}zetapars{n}pars{-,half}^{n - 1} -
sum_{n = 1}^{infty}bracks{zetapars{2n + 1} - 1}pars{half}^{2n} -
underbrace{sum_{n = 1}^{infty}pars{half}^{2n}}_{ds{1 over 3}}
\ = &
-bracks{Psipars{1 + z} + gamma}_{ z = -1/2}
\[3mm] & - bracks{%
{1 over 2z} - half,picotpars{pi z} - {1 over 1 - z^{2}} + 1 - gamma - Psipars{1 + z}}_{ z = 1/2} - {1 over 3}
\[8mm] = &
-Psipars{half} - {2 over 3} +
underbrace{Psipars{3 over 2}}_{ds{Psipars{1/2} + 1/pars{1/2}}} -
{1 over 3} = color{#f00}{1}
end{align}
$Psi$ and $gamma$ are the Digamma function and the Euler-Mascheroni constant, respectively. We used the Digamma Recurrence Formula $ds{Psipars{z + 1} = Psipars{z} + 1/z}$ and the identities $mathbf{6.3.14}$ and $mathbf{6.3.15}$ in this link.
answered Jun 19 '16 at 19:42
Felix Marin
67.1k7107141
$newcommand{angles}[1]{leftlangle{#1}rightrangle}
newcommand{braces}[1]{leftlbrace{#1}rightrbrace}
newcommand{bracks}[1]{leftlbrack{#1}rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Leftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left({#1}right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert{#1}rightvert}$
begin{align}
&bbox[10px,#ffd]{sum_{n = 1}^{infty}{zetapars{2n} over 2^{2n - 1}}} =
sum_{n = 2}^{infty}{zetapars{n} over 2^{n - 1}} - sum_{n = 1}^{infty}{zetapars{2n + 1} over 2^{2n}}
\[3mm] = &
-sum_{n = 2}^{infty}pars{-1}^{n}zetapars{n}pars{-,half}^{n - 1} -
sum_{n = 1}^{infty}bracks{zetapars{2n + 1} - 1}pars{half}^{2n} -
underbrace{sum_{n = 1}^{infty}pars{half}^{2n}}_{ds{1 over 3}}
\ = &
-bracks{Psipars{1 + z} + gamma}_{ z = -1/2}
\[3mm] & - bracks{%
{1 over 2z} - half,picotpars{pi z} - {1 over 1 - z^{2}} + 1 - gamma - Psipars{1 + z}}_{ z = 1/2} - {1 over 3}
\[8mm] = &
-Psipars{half} - {2 over 3} +
underbrace{Psipars{3 over 2}}_{ds{Psipars{1/2} + 1/pars{1/2}}} -
{1 over 3} = color{#f00}{1}
end{align}
$Psi$ and $gamma$ are the Digamma function and the Euler-Mascheroni constant, respectively. We used the Digamma Recurrence Formula $ds{Psipars{z + 1} = Psipars{z} + 1/z}$ and the identities $mathbf{6.3.14}$ and $mathbf{6.3.15}$ in this link.
answered Jun 19 '16 at 19:42
Felix Marin
67.1k7107141
edited Nov 19 '18 at 22:52
answered Jun 19 '16 at 19:42
Felix Marin
67.1k7107141
answered Jun 19 '16 at 19:42
Felix Marin
67.1k7107141
answered Jun 19 '16 at 19:42
Felix Marin
67.1k7107141
67.1k7107141
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1080000%2fhow-to-prove-that-frac-zeta2-2-frac-zeta-423-frac-zeta-62%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
12
I think this is a normal question. I don't know why "on hold"?
– E.H.E
Dec 24 '14 at 19:28
2
There is a recursive formula for $zeta(2n)$ derived in this answer. However, in that answer it is shown that $$sum_{k=1}^inftyzeta(2k)x^{2k}=frac12(1-pi xcot(pi x))$$ With $x=frac12$, this immediately gives $$sum_{k=1}^inftyzeta(2k)left(frac12right)^{2k-1} =1-fracpi2cotleft(fracpi2right)=1$$ This is what is used in Random Variable's answer.
– robjohn♦
Dec 31 '14 at 18:13