A nice relationship between $zeta$, $pi$ and $e$
I just happened to see this equation today, any suggestions on how to prove it?
$$sum_{n=1}^infty{frac{zeta(2n)}{n(2n+1)4^n}}=log{frac{pi}{e}}$$
real-analysis logarithms riemann-zeta pi eulers-constant
asked Dec 1 at 18:54
EvanHehehe
1247
add a comment |
I just happened to see this equation today, any suggestions on how to prove it?
$$sum_{n=1}^infty{frac{zeta(2n)}{n(2n+1)4^n}}=log{frac{pi}{e}}$$
real-analysis logarithms riemann-zeta pi eulers-constant
asked Dec 1 at 18:54
EvanHehehe
1247
Welcome to MSE. Have you attempted to prove it? If so, please include this in your post so that we can see you've made some effort.
– TheSimpliFire
Dec 1 at 18:56
1
Mmm, delicious...
– Frpzzd
Dec 1 at 19:03
Can you say where you saw this?
– Zacky
Dec 1 at 19:12
Okay, found the source: facebook.com/photo.php?fbid=1609500362528719. The second sum looks cute too: $$sum_{n=1}^infty left(frac{zeta(2n)}{n}-frac{zeta(2n+1)}{4^n}right)frac{1}{2n+1} =gamma +lnleft(frac{pi}{e}right)$$
– Zacky
Dec 1 at 19:36
@TheSimpliFire Thanks for the welcome. To be completely honest I have little experience in dealing with the zeta function and it's properties. I played around with the idea of using the Wallis product (as Frpzzd did in his answer) but ultimately gave up. I'll keep your suggestion in mind.
– EvanHehehe
Dec 1 at 21:10
add a comment |
I just happened to see this equation today, any suggestions on how to prove it?
$$sum_{n=1}^infty{frac{zeta(2n)}{n(2n+1)4^n}}=log{frac{pi}{e}}$$
real-analysis logarithms riemann-zeta pi eulers-constant
asked Dec 1 at 18:54
EvanHehehe
1247
I just happened to see this equation today, any suggestions on how to prove it?
$$sum_{n=1}^infty{frac{zeta(2n)}{n(2n+1)4^n}}=log{frac{pi}{e}}$$
real-analysis logarithms riemann-zeta pi eulers-constant
real-analysis logarithms riemann-zeta pi eulers-constant
asked Dec 1 at 18:54
EvanHehehe
1247
asked Dec 1 at 18:54
EvanHehehe
1247
asked Dec 1 at 18:54
EvanHehehe
1247
asked Dec 1 at 18:54
EvanHehehe
1247
asked Dec 1 at 18:54
EvanHehehe
1247
1247
Welcome to MSE. Have you attempted to prove it? If so, please include this in your post so that we can see you've made some effort.
– TheSimpliFire
Dec 1 at 18:56
1
Mmm, delicious...
– Frpzzd
Dec 1 at 19:03
Can you say where you saw this?
– Zacky
Dec 1 at 19:12
Okay, found the source: facebook.com/photo.php?fbid=1609500362528719. The second sum looks cute too: $$sum_{n=1}^infty left(frac{zeta(2n)}{n}-frac{zeta(2n+1)}{4^n}right)frac{1}{2n+1} =gamma +lnleft(frac{pi}{e}right)$$
– Zacky
Dec 1 at 19:36
@TheSimpliFire Thanks for the welcome. To be completely honest I have little experience in dealing with the zeta function and it's properties. I played around with the idea of using the Wallis product (as Frpzzd did in his answer) but ultimately gave up. I'll keep your suggestion in mind.
– EvanHehehe
Dec 1 at 21:10
add a comment |
Welcome to MSE. Have you attempted to prove it? If so, please include this in your post so that we can see you've made some effort.
– TheSimpliFire
Dec 1 at 18:56
1
Mmm, delicious...
– Frpzzd
Dec 1 at 19:03
Can you say where you saw this?
– Zacky
Dec 1 at 19:12
Okay, found the source: facebook.com/photo.php?fbid=1609500362528719. The second sum looks cute too: $$sum_{n=1}^infty left(frac{zeta(2n)}{n}-frac{zeta(2n+1)}{4^n}right)frac{1}{2n+1} =gamma +lnleft(frac{pi}{e}right)$$
– Zacky
Dec 1 at 19:36
@TheSimpliFire Thanks for the welcome. To be completely honest I have little experience in dealing with the zeta function and it's properties. I played around with the idea of using the Wallis product (as Frpzzd did in his answer) but ultimately gave up. I'll keep your suggestion in mind.
– EvanHehehe
Dec 1 at 21:10
Welcome to MSE. Have you attempted to prove it? If so, please include this in your post so that we can see you've made some effort.
– TheSimpliFire
Dec 1 at 18:56
Welcome to MSE. Have you attempted to prove it? If so, please include this in your post so that we can see you've made some effort.
– TheSimpliFire
Dec 1 at 18:56
1
1
Mmm, delicious...
– Frpzzd
Dec 1 at 19:03
Mmm, delicious...
– Frpzzd
Dec 1 at 19:03
Can you say where you saw this?
– Zacky
Dec 1 at 19:12
Can you say where you saw this?
– Zacky
Dec 1 at 19:12
Okay, found the source: facebook.com/photo.php?fbid=1609500362528719. The second sum looks cute too: $$sum_{n=1}^infty left(frac{zeta(2n)}{n}-frac{zeta(2n+1)}{4^n}right)frac{1}{2n+1} =gamma +lnleft(frac{pi}{e}right)$$
– Zacky
Dec 1 at 19:36
Okay, found the source: facebook.com/photo.php?fbid=1609500362528719. The second sum looks cute too: $$sum_{n=1}^infty left(frac{zeta(2n)}{n}-frac{zeta(2n+1)}{4^n}right)frac{1}{2n+1} =gamma +lnleft(frac{pi}{e}right)$$
– Zacky
Dec 1 at 19:36
@TheSimpliFire Thanks for the welcome. To be completely honest I have little experience in dealing with the zeta function and it's properties. I played around with the idea of using the Wallis product (as Frpzzd did in his answer) but ultimately gave up. I'll keep your suggestion in mind.
– EvanHehehe
Dec 1 at 21:10
@TheSimpliFire Thanks for the welcome. To be completely honest I have little experience in dealing with the zeta function and it's properties. I played around with the idea of using the Wallis product (as Frpzzd did in his answer) but ultimately gave up. I'll keep your suggestion in mind.
– EvanHehehe
Dec 1 at 21:10
add a comment |
4 Answers
4
active
oldest
votes
An approach that does not require complex analysis:
$$
sum_{n=1}^infty frac{zeta(2n)}{n(2n+1)4^n}
=sum_{n=1}^infty frac{zeta(2n)}{n4^n}-2sum_{n=1}^infty frac{zeta(2n)}{(2n+1)4^n}=S_1-2S_2
$$
To calculate $S_1$:
$$begin{align}
S_1=sum_{n=1}^infty frac{zeta(2n)}{n4^n}
&=sum_{n=1}^infty sum_{k=1}^infty frac{1}{n(4k^2)^n}\
&=sum_{k=1}^infty sum_{n=1}^infty frac{1}{n(4k^2)^n}\
&=-sum_{k=1}^infty lnbigg(1-frac{1}{4k^2}bigg)\
&=-ln prod_{k=1}^infty frac{(2k+1)(2k-1)}{(2k)^2}\
&=ln(pi/2)
end{align}$$
using the Wallis Product.
You may calculate $S_2$ similarly.
answered Dec 1 at 19:11
Frpzzd
21.8k839107
Thanks for this answer. I had the Wallis product in my mind but wasn't able to transform the equation into anything coherent.
– EvanHehehe
Dec 1 at 21:14
@EvanHehehe No problem, I'm glad to help.
– Frpzzd
Dec 1 at 21:15
add a comment |
The left-hand side is $$sum_{nge 1}frac{1}{Gamma (2n)n(2n+1)4^n}int_0^inftyfrac{x^{2n-1} dx}{e^x-1}=int_0^inftyfrac{dx}{e^x-1}sum_{nge 1}frac{(x/2)^{2n-1}}{(2n+1)!}\=int_0^inftyfrac{dx}{e^x-1}frac{sinhtfrac{x}{2}-tfrac{x}{2}}{(tfrac{x}{2})^2}.$$The rest is an exercise in complex analysis.
answered Dec 1 at 19:03
J.G.
22.3k22035
add a comment |
The identities
$$begin{align}sum_{k=1}^{infty}frac{zetaleft(2kright)}{k}z^{2k}&=lnleft(frac{pi z}{sinleft(pi zright)}right)&&(lvert z rvert < 1)\
sum_{k=1}^{infty}frac{zetaleft(2kright)}{(2k+1)2^{2k}}&=frac{1}{2}-frac
{1}{2}ln 2end{align}$$
(DLMF 25.8.8, 25.8.9) follow from the more basic identity
$$begin{align}
sum_{k=2}^{infty}frac{zetaleft(kright)}{k}z^{k}&=-gamma z+lnGammaleft
(1-zright)&&(lvert z rvert < 1)
end{align}$$
(DLMF 25.8.7), which in turn follows from the product formula for the Gamma function (DLMF 5.8.2):
$$frac{1}{Gammaleft(zright)}=ze^{gamma z}prod_{k=1}^{infty}left(1+frac{%
z}{k}right)mathrm{e}^{-z/k}text{.}$$
answered Dec 1 at 19:52
K B Dave
3,252217
you meant looking at $sum_{k=2}^{infty}frac{zetaleft(kright)}{k+1}z^{k+1}= ?$
– reuns
Dec 1 at 20:07
add a comment |
Since
begin{align*}
zeta(2n) &= frac{(-1)^{n+1} B_{2n} (2pi)^{2n}}{2(2n)!},
end{align*}
for integers $n > 0$, the given sum $S$ is
begin{align*}
S &= sum_{n=1}^infty frac{zeta(2n)}{n(2n+1)4^n}
= sum_{n=1}^infty (-1)^{n+1} frac{B_{2n}}{(2n)(2n+1)}frac{pi^{2n}}{(2n)!}
end{align*}
But
begin{align*}
f(z) &= sum_{n=1}^infty frac{B_n}{n(n+1)} frac{z^{n+1}}{n!}
end{align*}
has $operatorname{Im} f(pi i) = pi S$ and
begin{align*}
f''(z) &= frac{1}{z}sum_{n=1}^infty B_n frac{z^n}{n!} = frac{1}{z}left(-1 + frac{z}{e^z - 1}right) = -frac{1}{z} + frac{1}{e^z - 1}
end{align*}
by (a) definition of the Bernoulli numbers $B_n$. A bit of careful integration gives the required sum.
answered Dec 1 at 19:50
anomaly
17.3k42663
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
An approach that does not require complex analysis:
$$
sum_{n=1}^infty frac{zeta(2n)}{n(2n+1)4^n}
=sum_{n=1}^infty frac{zeta(2n)}{n4^n}-2sum_{n=1}^infty frac{zeta(2n)}{(2n+1)4^n}=S_1-2S_2
$$
To calculate $S_1$:
$$begin{align}
S_1=sum_{n=1}^infty frac{zeta(2n)}{n4^n}
&=sum_{n=1}^infty sum_{k=1}^infty frac{1}{n(4k^2)^n}\
&=sum_{k=1}^infty sum_{n=1}^infty frac{1}{n(4k^2)^n}\
&=-sum_{k=1}^infty lnbigg(1-frac{1}{4k^2}bigg)\
&=-ln prod_{k=1}^infty frac{(2k+1)(2k-1)}{(2k)^2}\
&=ln(pi/2)
end{align}$$
using the Wallis Product.
You may calculate $S_2$ similarly.
answered Dec 1 at 19:11
Frpzzd
21.8k839107
Thanks for this answer. I had the Wallis product in my mind but wasn't able to transform the equation into anything coherent.
– EvanHehehe
Dec 1 at 21:14
@EvanHehehe No problem, I'm glad to help.
– Frpzzd
Dec 1 at 21:15
add a comment |
An approach that does not require complex analysis:
$$
sum_{n=1}^infty frac{zeta(2n)}{n(2n+1)4^n}
=sum_{n=1}^infty frac{zeta(2n)}{n4^n}-2sum_{n=1}^infty frac{zeta(2n)}{(2n+1)4^n}=S_1-2S_2
$$
To calculate $S_1$:
$$begin{align}
S_1=sum_{n=1}^infty frac{zeta(2n)}{n4^n}
&=sum_{n=1}^infty sum_{k=1}^infty frac{1}{n(4k^2)^n}\
&=sum_{k=1}^infty sum_{n=1}^infty frac{1}{n(4k^2)^n}\
&=-sum_{k=1}^infty lnbigg(1-frac{1}{4k^2}bigg)\
&=-ln prod_{k=1}^infty frac{(2k+1)(2k-1)}{(2k)^2}\
&=ln(pi/2)
end{align}$$
using the Wallis Product.
You may calculate $S_2$ similarly.
answered Dec 1 at 19:11
Frpzzd
21.8k839107
Thanks for this answer. I had the Wallis product in my mind but wasn't able to transform the equation into anything coherent.
– EvanHehehe
Dec 1 at 21:14
@EvanHehehe No problem, I'm glad to help.
– Frpzzd
Dec 1 at 21:15
add a comment |
An approach that does not require complex analysis:
$$
sum_{n=1}^infty frac{zeta(2n)}{n(2n+1)4^n}
=sum_{n=1}^infty frac{zeta(2n)}{n4^n}-2sum_{n=1}^infty frac{zeta(2n)}{(2n+1)4^n}=S_1-2S_2
$$
To calculate $S_1$:
$$begin{align}
S_1=sum_{n=1}^infty frac{zeta(2n)}{n4^n}
&=sum_{n=1}^infty sum_{k=1}^infty frac{1}{n(4k^2)^n}\
&=sum_{k=1}^infty sum_{n=1}^infty frac{1}{n(4k^2)^n}\
&=-sum_{k=1}^infty lnbigg(1-frac{1}{4k^2}bigg)\
&=-ln prod_{k=1}^infty frac{(2k+1)(2k-1)}{(2k)^2}\
&=ln(pi/2)
end{align}$$
using the Wallis Product.
You may calculate $S_2$ similarly.
answered Dec 1 at 19:11
Frpzzd
21.8k839107
An approach that does not require complex analysis:
$$
sum_{n=1}^infty frac{zeta(2n)}{n(2n+1)4^n}
=sum_{n=1}^infty frac{zeta(2n)}{n4^n}-2sum_{n=1}^infty frac{zeta(2n)}{(2n+1)4^n}=S_1-2S_2
$$
To calculate $S_1$:
$$begin{align}
S_1=sum_{n=1}^infty frac{zeta(2n)}{n4^n}
&=sum_{n=1}^infty sum_{k=1}^infty frac{1}{n(4k^2)^n}\
&=sum_{k=1}^infty sum_{n=1}^infty frac{1}{n(4k^2)^n}\
&=-sum_{k=1}^infty lnbigg(1-frac{1}{4k^2}bigg)\
&=-ln prod_{k=1}^infty frac{(2k+1)(2k-1)}{(2k)^2}\
&=ln(pi/2)
end{align}$$
using the Wallis Product.
You may calculate $S_2$ similarly.
answered Dec 1 at 19:11
Frpzzd
21.8k839107
edited Dec 1 at 19:31
answered Dec 1 at 19:11
Frpzzd
21.8k839107
answered Dec 1 at 19:11
Frpzzd
21.8k839107
answered Dec 1 at 19:11
Frpzzd
21.8k839107
21.8k839107
Thanks for this answer. I had the Wallis product in my mind but wasn't able to transform the equation into anything coherent.
– EvanHehehe
Dec 1 at 21:14
@EvanHehehe No problem, I'm glad to help.
– Frpzzd
Dec 1 at 21:15
add a comment |
Thanks for this answer. I had the Wallis product in my mind but wasn't able to transform the equation into anything coherent.
– EvanHehehe
Dec 1 at 21:14
@EvanHehehe No problem, I'm glad to help.
– Frpzzd
Dec 1 at 21:15
Thanks for this answer. I had the Wallis product in my mind but wasn't able to transform the equation into anything coherent.
– EvanHehehe
Dec 1 at 21:14
Thanks for this answer. I had the Wallis product in my mind but wasn't able to transform the equation into anything coherent.
– EvanHehehe
Dec 1 at 21:14
@EvanHehehe No problem, I'm glad to help.
– Frpzzd
Dec 1 at 21:15
@EvanHehehe No problem, I'm glad to help.
– Frpzzd
Dec 1 at 21:15
add a comment |
The left-hand side is $$sum_{nge 1}frac{1}{Gamma (2n)n(2n+1)4^n}int_0^inftyfrac{x^{2n-1} dx}{e^x-1}=int_0^inftyfrac{dx}{e^x-1}sum_{nge 1}frac{(x/2)^{2n-1}}{(2n+1)!}\=int_0^inftyfrac{dx}{e^x-1}frac{sinhtfrac{x}{2}-tfrac{x}{2}}{(tfrac{x}{2})^2}.$$The rest is an exercise in complex analysis.
answered Dec 1 at 19:03
J.G.
22.3k22035
add a comment |
The left-hand side is $$sum_{nge 1}frac{1}{Gamma (2n)n(2n+1)4^n}int_0^inftyfrac{x^{2n-1} dx}{e^x-1}=int_0^inftyfrac{dx}{e^x-1}sum_{nge 1}frac{(x/2)^{2n-1}}{(2n+1)!}\=int_0^inftyfrac{dx}{e^x-1}frac{sinhtfrac{x}{2}-tfrac{x}{2}}{(tfrac{x}{2})^2}.$$The rest is an exercise in complex analysis.
answered Dec 1 at 19:03
J.G.
22.3k22035
add a comment |
The left-hand side is $$sum_{nge 1}frac{1}{Gamma (2n)n(2n+1)4^n}int_0^inftyfrac{x^{2n-1} dx}{e^x-1}=int_0^inftyfrac{dx}{e^x-1}sum_{nge 1}frac{(x/2)^{2n-1}}{(2n+1)!}\=int_0^inftyfrac{dx}{e^x-1}frac{sinhtfrac{x}{2}-tfrac{x}{2}}{(tfrac{x}{2})^2}.$$The rest is an exercise in complex analysis.
answered Dec 1 at 19:03
J.G.
22.3k22035
The left-hand side is $$sum_{nge 1}frac{1}{Gamma (2n)n(2n+1)4^n}int_0^inftyfrac{x^{2n-1} dx}{e^x-1}=int_0^inftyfrac{dx}{e^x-1}sum_{nge 1}frac{(x/2)^{2n-1}}{(2n+1)!}\=int_0^inftyfrac{dx}{e^x-1}frac{sinhtfrac{x}{2}-tfrac{x}{2}}{(tfrac{x}{2})^2}.$$The rest is an exercise in complex analysis.
answered Dec 1 at 19:03
J.G.
22.3k22035
edited Dec 1 at 19:25
answered Dec 1 at 19:03
J.G.
22.3k22035
answered Dec 1 at 19:03
J.G.
22.3k22035
answered Dec 1 at 19:03
J.G.
22.3k22035
22.3k22035
add a comment |
add a comment |
The identities
$$begin{align}sum_{k=1}^{infty}frac{zetaleft(2kright)}{k}z^{2k}&=lnleft(frac{pi z}{sinleft(pi zright)}right)&&(lvert z rvert < 1)\
sum_{k=1}^{infty}frac{zetaleft(2kright)}{(2k+1)2^{2k}}&=frac{1}{2}-frac
{1}{2}ln 2end{align}$$
(DLMF 25.8.8, 25.8.9) follow from the more basic identity
$$begin{align}
sum_{k=2}^{infty}frac{zetaleft(kright)}{k}z^{k}&=-gamma z+lnGammaleft
(1-zright)&&(lvert z rvert < 1)
end{align}$$
(DLMF 25.8.7), which in turn follows from the product formula for the Gamma function (DLMF 5.8.2):
$$frac{1}{Gammaleft(zright)}=ze^{gamma z}prod_{k=1}^{infty}left(1+frac{%
z}{k}right)mathrm{e}^{-z/k}text{.}$$
answered Dec 1 at 19:52
K B Dave
3,252217
you meant looking at $sum_{k=2}^{infty}frac{zetaleft(kright)}{k+1}z^{k+1}= ?$
– reuns
Dec 1 at 20:07
add a comment |
The identities
$$begin{align}sum_{k=1}^{infty}frac{zetaleft(2kright)}{k}z^{2k}&=lnleft(frac{pi z}{sinleft(pi zright)}right)&&(lvert z rvert < 1)\
sum_{k=1}^{infty}frac{zetaleft(2kright)}{(2k+1)2^{2k}}&=frac{1}{2}-frac
{1}{2}ln 2end{align}$$
(DLMF 25.8.8, 25.8.9) follow from the more basic identity
$$begin{align}
sum_{k=2}^{infty}frac{zetaleft(kright)}{k}z^{k}&=-gamma z+lnGammaleft
(1-zright)&&(lvert z rvert < 1)
end{align}$$
(DLMF 25.8.7), which in turn follows from the product formula for the Gamma function (DLMF 5.8.2):
$$frac{1}{Gammaleft(zright)}=ze^{gamma z}prod_{k=1}^{infty}left(1+frac{%
z}{k}right)mathrm{e}^{-z/k}text{.}$$
answered Dec 1 at 19:52
K B Dave
3,252217
you meant looking at $sum_{k=2}^{infty}frac{zetaleft(kright)}{k+1}z^{k+1}= ?$
– reuns
Dec 1 at 20:07
add a comment |
The identities
$$begin{align}sum_{k=1}^{infty}frac{zetaleft(2kright)}{k}z^{2k}&=lnleft(frac{pi z}{sinleft(pi zright)}right)&&(lvert z rvert < 1)\
sum_{k=1}^{infty}frac{zetaleft(2kright)}{(2k+1)2^{2k}}&=frac{1}{2}-frac
{1}{2}ln 2end{align}$$
(DLMF 25.8.8, 25.8.9) follow from the more basic identity
$$begin{align}
sum_{k=2}^{infty}frac{zetaleft(kright)}{k}z^{k}&=-gamma z+lnGammaleft
(1-zright)&&(lvert z rvert < 1)
end{align}$$
(DLMF 25.8.7), which in turn follows from the product formula for the Gamma function (DLMF 5.8.2):
$$frac{1}{Gammaleft(zright)}=ze^{gamma z}prod_{k=1}^{infty}left(1+frac{%
z}{k}right)mathrm{e}^{-z/k}text{.}$$
answered Dec 1 at 19:52
K B Dave
3,252217
The identities
$$begin{align}sum_{k=1}^{infty}frac{zetaleft(2kright)}{k}z^{2k}&=lnleft(frac{pi z}{sinleft(pi zright)}right)&&(lvert z rvert < 1)\
sum_{k=1}^{infty}frac{zetaleft(2kright)}{(2k+1)2^{2k}}&=frac{1}{2}-frac
{1}{2}ln 2end{align}$$
(DLMF 25.8.8, 25.8.9) follow from the more basic identity
$$begin{align}
sum_{k=2}^{infty}frac{zetaleft(kright)}{k}z^{k}&=-gamma z+lnGammaleft
(1-zright)&&(lvert z rvert < 1)
end{align}$$
(DLMF 25.8.7), which in turn follows from the product formula for the Gamma function (DLMF 5.8.2):
$$frac{1}{Gammaleft(zright)}=ze^{gamma z}prod_{k=1}^{infty}left(1+frac{%
z}{k}right)mathrm{e}^{-z/k}text{.}$$
answered Dec 1 at 19:52
K B Dave
3,252217
answered Dec 1 at 19:52
K B Dave
3,252217
answered Dec 1 at 19:52
K B Dave
3,252217
answered Dec 1 at 19:52
K B Dave
3,252217
3,252217
you meant looking at $sum_{k=2}^{infty}frac{zetaleft(kright)}{k+1}z^{k+1}= ?$
– reuns
Dec 1 at 20:07
add a comment |
you meant looking at $sum_{k=2}^{infty}frac{zetaleft(kright)}{k+1}z^{k+1}= ?$
– reuns
Dec 1 at 20:07
you meant looking at $sum_{k=2}^{infty}frac{zetaleft(kright)}{k+1}z^{k+1}= ?$
– reuns
Dec 1 at 20:07
you meant looking at $sum_{k=2}^{infty}frac{zetaleft(kright)}{k+1}z^{k+1}= ?$
– reuns
Dec 1 at 20:07
add a comment |
Since
begin{align*}
zeta(2n) &= frac{(-1)^{n+1} B_{2n} (2pi)^{2n}}{2(2n)!},
end{align*}
for integers $n > 0$, the given sum $S$ is
begin{align*}
S &= sum_{n=1}^infty frac{zeta(2n)}{n(2n+1)4^n}
= sum_{n=1}^infty (-1)^{n+1} frac{B_{2n}}{(2n)(2n+1)}frac{pi^{2n}}{(2n)!}
end{align*}
But
begin{align*}
f(z) &= sum_{n=1}^infty frac{B_n}{n(n+1)} frac{z^{n+1}}{n!}
end{align*}
has $operatorname{Im} f(pi i) = pi S$ and
begin{align*}
f''(z) &= frac{1}{z}sum_{n=1}^infty B_n frac{z^n}{n!} = frac{1}{z}left(-1 + frac{z}{e^z - 1}right) = -frac{1}{z} + frac{1}{e^z - 1}
end{align*}
by (a) definition of the Bernoulli numbers $B_n$. A bit of careful integration gives the required sum.
answered Dec 1 at 19:50
anomaly
17.3k42663
add a comment |
Since
begin{align*}
zeta(2n) &= frac{(-1)^{n+1} B_{2n} (2pi)^{2n}}{2(2n)!},
end{align*}
for integers $n > 0$, the given sum $S$ is
begin{align*}
S &= sum_{n=1}^infty frac{zeta(2n)}{n(2n+1)4^n}
= sum_{n=1}^infty (-1)^{n+1} frac{B_{2n}}{(2n)(2n+1)}frac{pi^{2n}}{(2n)!}
end{align*}
But
begin{align*}
f(z) &= sum_{n=1}^infty frac{B_n}{n(n+1)} frac{z^{n+1}}{n!}
end{align*}
has $operatorname{Im} f(pi i) = pi S$ and
begin{align*}
f''(z) &= frac{1}{z}sum_{n=1}^infty B_n frac{z^n}{n!} = frac{1}{z}left(-1 + frac{z}{e^z - 1}right) = -frac{1}{z} + frac{1}{e^z - 1}
end{align*}
by (a) definition of the Bernoulli numbers $B_n$. A bit of careful integration gives the required sum.
answered Dec 1 at 19:50
anomaly
17.3k42663
add a comment |
Since
begin{align*}
zeta(2n) &= frac{(-1)^{n+1} B_{2n} (2pi)^{2n}}{2(2n)!},
end{align*}
for integers $n > 0$, the given sum $S$ is
begin{align*}
S &= sum_{n=1}^infty frac{zeta(2n)}{n(2n+1)4^n}
= sum_{n=1}^infty (-1)^{n+1} frac{B_{2n}}{(2n)(2n+1)}frac{pi^{2n}}{(2n)!}
end{align*}
But
begin{align*}
f(z) &= sum_{n=1}^infty frac{B_n}{n(n+1)} frac{z^{n+1}}{n!}
end{align*}
has $operatorname{Im} f(pi i) = pi S$ and
begin{align*}
f''(z) &= frac{1}{z}sum_{n=1}^infty B_n frac{z^n}{n!} = frac{1}{z}left(-1 + frac{z}{e^z - 1}right) = -frac{1}{z} + frac{1}{e^z - 1}
end{align*}
by (a) definition of the Bernoulli numbers $B_n$. A bit of careful integration gives the required sum.
answered Dec 1 at 19:50
anomaly
17.3k42663
Since
begin{align*}
zeta(2n) &= frac{(-1)^{n+1} B_{2n} (2pi)^{2n}}{2(2n)!},
end{align*}
for integers $n > 0$, the given sum $S$ is
begin{align*}
S &= sum_{n=1}^infty frac{zeta(2n)}{n(2n+1)4^n}
= sum_{n=1}^infty (-1)^{n+1} frac{B_{2n}}{(2n)(2n+1)}frac{pi^{2n}}{(2n)!}
end{align*}
But
begin{align*}
f(z) &= sum_{n=1}^infty frac{B_n}{n(n+1)} frac{z^{n+1}}{n!}
end{align*}
has $operatorname{Im} f(pi i) = pi S$ and
begin{align*}
f''(z) &= frac{1}{z}sum_{n=1}^infty B_n frac{z^n}{n!} = frac{1}{z}left(-1 + frac{z}{e^z - 1}right) = -frac{1}{z} + frac{1}{e^z - 1}
end{align*}
by (a) definition of the Bernoulli numbers $B_n$. A bit of careful integration gives the required sum.
answered Dec 1 at 19:50
anomaly
17.3k42663
answered Dec 1 at 19:50
anomaly
17.3k42663
answered Dec 1 at 19:50
anomaly
17.3k42663
answered Dec 1 at 19:50
anomaly
17.3k42663
17.3k42663
add a comment |
add a comment |
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Welcome to MSE. Have you attempted to prove it? If so, please include this in your post so that we can see you've made some effort.
– TheSimpliFire
Dec 1 at 18:56
1
Mmm, delicious...
– Frpzzd
Dec 1 at 19:03
Can you say where you saw this?
– Zacky
Dec 1 at 19:12
Okay, found the source: facebook.com/photo.php?fbid=1609500362528719. The second sum looks cute too: $$sum_{n=1}^infty left(frac{zeta(2n)}{n}-frac{zeta(2n+1)}{4^n}right)frac{1}{2n+1} =gamma +lnleft(frac{pi}{e}right)$$
– Zacky
Dec 1 at 19:36
@TheSimpliFire Thanks for the welcome. To be completely honest I have little experience in dealing with the zeta function and it's properties. I played around with the idea of using the Wallis product (as Frpzzd did in his answer) but ultimately gave up. I'll keep your suggestion in mind.
– EvanHehehe
Dec 1 at 21:10