Proof for this identity: $sumlimits_k=0^infty(zeta(2k+3)-1) = frac14$
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
$$
sum_k=0^infty (zeta(2k+3)
-1) = frac14
$$
I found this nice identity and want to prove it, but I don't know how.
My first thought was that I can create a geometric series with the same limit like:
$$sum_k=0^infty frac18 cdot frac12^k = frac14$$
But I have no idea, how to write the first one in a way, that these are the same. They are obviously only equal, if we sum infinitely many terms together.
Do you have ideas, how to proof this identity or do you know, where the proof can be found?
Thank you for your answers!
sequences-and-series riemann-zeta geometric-series
edited Mar 30 at 16:32
Marco Cantarini
28.8k23272
asked Mar 30 at 13:37
Mister Set
617310
add a comment |
up vote
2
down vote
favorite
$$
sum_k=0^infty (zeta(2k+3)
-1) = frac14
$$
I found this nice identity and want to prove it, but I don't know how.
My first thought was that I can create a geometric series with the same limit like:
$$sum_k=0^infty frac18 cdot frac12^k = frac14$$
But I have no idea, how to write the first one in a way, that these are the same. They are obviously only equal, if we sum infinitely many terms together.
Do you have ideas, how to proof this identity or do you know, where the proof can be found?
Thank you for your answers!
sequences-and-series riemann-zeta geometric-series
edited Mar 30 at 16:32
Marco Cantarini
28.8k23272
asked Mar 30 at 13:37
Mister Set
617310
2
Where did you found it? All terms are greater than 1!
– gammatester
Mar 30 at 13:41
2
You have a typo, should be $sum_k=0^infty (zeta(2k+3) - colorred1) = frac14$.
– achille hui
Mar 30 at 13:50
Actually , this sum seems to be $k+frac54$ with a very small error already for $k=10$
– Peter
Mar 30 at 13:51
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$$
sum_k=0^infty (zeta(2k+3)
-1) = frac14
$$
I found this nice identity and want to prove it, but I don't know how.
My first thought was that I can create a geometric series with the same limit like:
$$sum_k=0^infty frac18 cdot frac12^k = frac14$$
But I have no idea, how to write the first one in a way, that these are the same. They are obviously only equal, if we sum infinitely many terms together.
Do you have ideas, how to proof this identity or do you know, where the proof can be found?
Thank you for your answers!
sequences-and-series riemann-zeta geometric-series
edited Mar 30 at 16:32
Marco Cantarini
28.8k23272
asked Mar 30 at 13:37
Mister Set
617310
$$
sum_k=0^infty (zeta(2k+3)
-1) = frac14
$$
I found this nice identity and want to prove it, but I don't know how.
My first thought was that I can create a geometric series with the same limit like:
$$sum_k=0^infty frac18 cdot frac12^k = frac14$$
But I have no idea, how to write the first one in a way, that these are the same. They are obviously only equal, if we sum infinitely many terms together.
Do you have ideas, how to proof this identity or do you know, where the proof can be found?
Thank you for your answers!
sequences-and-series riemann-zeta geometric-series
sequences-and-series riemann-zeta geometric-series
edited Mar 30 at 16:32
Marco Cantarini
28.8k23272
asked Mar 30 at 13:37
Mister Set
617310
edited Mar 30 at 16:32
Marco Cantarini
28.8k23272
asked Mar 30 at 13:37
Mister Set
617310
edited Mar 30 at 16:32
Marco Cantarini
28.8k23272
edited Mar 30 at 16:32
Marco Cantarini
28.8k23272
edited Mar 30 at 16:32
Marco Cantarini
28.8k23272
28.8k23272
asked Mar 30 at 13:37
Mister Set
617310
asked Mar 30 at 13:37
Mister Set
617310
asked Mar 30 at 13:37
Mister Set
617310
617310
2
Where did you found it? All terms are greater than 1!
– gammatester
Mar 30 at 13:41
2
You have a typo, should be $sum_k=0^infty (zeta(2k+3) - colorred1) = frac14$.
– achille hui
Mar 30 at 13:50
Actually , this sum seems to be $k+frac54$ with a very small error already for $k=10$
– Peter
Mar 30 at 13:51
add a comment |
2
Where did you found it? All terms are greater than 1!
– gammatester
Mar 30 at 13:41
2
You have a typo, should be $sum_k=0^infty (zeta(2k+3) - colorred1) = frac14$.
– achille hui
Mar 30 at 13:50
Actually , this sum seems to be $k+frac54$ with a very small error already for $k=10$
– Peter
Mar 30 at 13:51
2
2
Where did you found it? All terms are greater than 1!
– gammatester
Mar 30 at 13:41
Where did you found it? All terms are greater than 1!
– gammatester
Mar 30 at 13:41
2
2
You have a typo, should be $sum_k=0^infty (zeta(2k+3) - colorred1) = frac14$.
– achille hui
Mar 30 at 13:50
You have a typo, should be $sum_k=0^infty (zeta(2k+3) - colorred1) = frac14$.
– achille hui
Mar 30 at 13:50
Actually , this sum seems to be $k+frac54$ with a very small error already for $k=10$
– Peter
Mar 30 at 13:51
Actually , this sum seems to be $k+frac54$ with a very small error already for $k=10$
– Peter
Mar 30 at 13:51
add a comment |
3 Answers
3
active
oldest
votes
up vote
5
down vote
There is a typo in the question.
The sum $sumlimits_k=0^infty zeta(2k+3)$ diverges because $limlimits_ktoinfty zeta(2k+3) = 1$.
The correct statement should be
$$sum_k=0^infty (zeta(2k+3) - 1) = frac14$$
The proof is relatively simple,
$$beginalign sum_k=0^infty (zeta(2k+3)-1)
= &sum_k=0^infty sum_n=2^infty frac1n^2k+3
stackrelcolorblue[1]= sum_n=2^infty sum_k=0^infty frac1n^2k+3
= sum_n=2^infty frac1n^3sum_k=0^infty (n^-2)^k\
= &sum_n=2^infty frac1n^3frac11 - n^-2
= sum_n=2^infty frac1(n-1)n(n+1)
= sum_n=2^infty frac12left(frac1(n-1)n - frac1n(n+1)right)\
stackrelcolorblue[2]= & frac12left(frac1(2-1)2right) = frac14endalign
$$
Notes
$colorblue[1]$ - all terms are non-negative, it is legal to exchange order of summation.
$colorblue[2]$ - the sum is a telescoping one.
answered Mar 30 at 14:12
achille hui
92.8k5127249
Wow, so easy is that (if one has the "know-how") (+1)
– Peter
Mar 30 at 14:16
The concept of a telescoping sum is very fascinating. I was not aware of this type of series
– Mister Set
Mar 30 at 17:15
add a comment |
up vote
3
down vote
By the integral representation of $zetaleft(sright)$ we have $$zetaleft(2k+3right)-1=frac1left(2k+2right)!int_0^inftyfracu^2k+2e^uleft(e^u-1right)du$$ then $$sum_kgeq0left(zetaleft(2k+3right)-1right)=int_0^inftyfrac1e^uleft(e^u-1right)sum_kgeq0fracu^2k+2left(2k+2right)!du$$ $$=int_0^inftyfraccoshleft(uright)-1e^uleft(e^u-1right)du=frac12int_0^inftye^-u-e^-2udu=color redfrac14.$$
answered Mar 30 at 15:51
Marco Cantarini
28.8k23272
add a comment |
up vote
0
down vote
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandroot[2],sqrt[#1],#2,,
newcommandtotald[3]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
&bbox[10px,#ffd]dssum_k = 0^inftybrackszetapars2k + 3 - 1 =
sum_k = 1^inftybrackszetapars2k + 1 - 1
\[5mm] = &
lim_z to 1bracks1 over 2z - 1 over 2,pi,cotparspi z -
1 over 1 - z^2 + 1 - gamma - Psipars1 + z = bbx1 over 4
endalign
where $dsPsi$ is the Digamma Function. See $dsmathbfcolorblack6.3.15$ in A & S Table.
Note that
$dsPsipars2 = Psipars1 + 1 = -gamma + 1$.
answered Apr 2 at 3:41
Felix Marin
65.6k7107138
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
There is a typo in the question.
The sum $sumlimits_k=0^infty zeta(2k+3)$ diverges because $limlimits_ktoinfty zeta(2k+3) = 1$.
The correct statement should be
$$sum_k=0^infty (zeta(2k+3) - 1) = frac14$$
The proof is relatively simple,
$$beginalign sum_k=0^infty (zeta(2k+3)-1)
= &sum_k=0^infty sum_n=2^infty frac1n^2k+3
stackrelcolorblue[1]= sum_n=2^infty sum_k=0^infty frac1n^2k+3
= sum_n=2^infty frac1n^3sum_k=0^infty (n^-2)^k\
= &sum_n=2^infty frac1n^3frac11 - n^-2
= sum_n=2^infty frac1(n-1)n(n+1)
= sum_n=2^infty frac12left(frac1(n-1)n - frac1n(n+1)right)\
stackrelcolorblue[2]= & frac12left(frac1(2-1)2right) = frac14endalign
$$
Notes
$colorblue[1]$ - all terms are non-negative, it is legal to exchange order of summation.
$colorblue[2]$ - the sum is a telescoping one.
answered Mar 30 at 14:12
achille hui
92.8k5127249
Wow, so easy is that (if one has the "know-how") (+1)
– Peter
Mar 30 at 14:16
The concept of a telescoping sum is very fascinating. I was not aware of this type of series
– Mister Set
Mar 30 at 17:15
add a comment |
up vote
5
down vote
There is a typo in the question.
The sum $sumlimits_k=0^infty zeta(2k+3)$ diverges because $limlimits_ktoinfty zeta(2k+3) = 1$.
The correct statement should be
$$sum_k=0^infty (zeta(2k+3) - 1) = frac14$$
The proof is relatively simple,
$$beginalign sum_k=0^infty (zeta(2k+3)-1)
= &sum_k=0^infty sum_n=2^infty frac1n^2k+3
stackrelcolorblue[1]= sum_n=2^infty sum_k=0^infty frac1n^2k+3
= sum_n=2^infty frac1n^3sum_k=0^infty (n^-2)^k\
= &sum_n=2^infty frac1n^3frac11 - n^-2
= sum_n=2^infty frac1(n-1)n(n+1)
= sum_n=2^infty frac12left(frac1(n-1)n - frac1n(n+1)right)\
stackrelcolorblue[2]= & frac12left(frac1(2-1)2right) = frac14endalign
$$
Notes
$colorblue[1]$ - all terms are non-negative, it is legal to exchange order of summation.
$colorblue[2]$ - the sum is a telescoping one.
answered Mar 30 at 14:12
achille hui
92.8k5127249
Wow, so easy is that (if one has the "know-how") (+1)
– Peter
Mar 30 at 14:16
The concept of a telescoping sum is very fascinating. I was not aware of this type of series
– Mister Set
Mar 30 at 17:15
add a comment |
up vote
5
down vote
up vote
5
down vote
There is a typo in the question.
The sum $sumlimits_k=0^infty zeta(2k+3)$ diverges because $limlimits_ktoinfty zeta(2k+3) = 1$.
The correct statement should be
$$sum_k=0^infty (zeta(2k+3) - 1) = frac14$$
The proof is relatively simple,
$$beginalign sum_k=0^infty (zeta(2k+3)-1)
= &sum_k=0^infty sum_n=2^infty frac1n^2k+3
stackrelcolorblue[1]= sum_n=2^infty sum_k=0^infty frac1n^2k+3
= sum_n=2^infty frac1n^3sum_k=0^infty (n^-2)^k\
= &sum_n=2^infty frac1n^3frac11 - n^-2
= sum_n=2^infty frac1(n-1)n(n+1)
= sum_n=2^infty frac12left(frac1(n-1)n - frac1n(n+1)right)\
stackrelcolorblue[2]= & frac12left(frac1(2-1)2right) = frac14endalign
$$
Notes
$colorblue[1]$ - all terms are non-negative, it is legal to exchange order of summation.
$colorblue[2]$ - the sum is a telescoping one.
answered Mar 30 at 14:12
achille hui
92.8k5127249
There is a typo in the question.
The sum $sumlimits_k=0^infty zeta(2k+3)$ diverges because $limlimits_ktoinfty zeta(2k+3) = 1$.
The correct statement should be
$$sum_k=0^infty (zeta(2k+3) - 1) = frac14$$
The proof is relatively simple,
$$beginalign sum_k=0^infty (zeta(2k+3)-1)
= &sum_k=0^infty sum_n=2^infty frac1n^2k+3
stackrelcolorblue[1]= sum_n=2^infty sum_k=0^infty frac1n^2k+3
= sum_n=2^infty frac1n^3sum_k=0^infty (n^-2)^k\
= &sum_n=2^infty frac1n^3frac11 - n^-2
= sum_n=2^infty frac1(n-1)n(n+1)
= sum_n=2^infty frac12left(frac1(n-1)n - frac1n(n+1)right)\
stackrelcolorblue[2]= & frac12left(frac1(2-1)2right) = frac14endalign
$$
Notes
$colorblue[1]$ - all terms are non-negative, it is legal to exchange order of summation.
$colorblue[2]$ - the sum is a telescoping one.
answered Mar 30 at 14:12
achille hui
92.8k5127249
answered Mar 30 at 14:12
achille hui
92.8k5127249
answered Mar 30 at 14:12
achille hui
92.8k5127249
answered Mar 30 at 14:12
achille hui
92.8k5127249
92.8k5127249
Wow, so easy is that (if one has the "know-how") (+1)
– Peter
Mar 30 at 14:16
The concept of a telescoping sum is very fascinating. I was not aware of this type of series
– Mister Set
Mar 30 at 17:15
add a comment |
Wow, so easy is that (if one has the "know-how") (+1)
– Peter
Mar 30 at 14:16
The concept of a telescoping sum is very fascinating. I was not aware of this type of series
– Mister Set
Mar 30 at 17:15
Wow, so easy is that (if one has the "know-how") (+1)
– Peter
Mar 30 at 14:16
Wow, so easy is that (if one has the "know-how") (+1)
– Peter
Mar 30 at 14:16
The concept of a telescoping sum is very fascinating. I was not aware of this type of series
– Mister Set
Mar 30 at 17:15
The concept of a telescoping sum is very fascinating. I was not aware of this type of series
– Mister Set
Mar 30 at 17:15
add a comment |
up vote
3
down vote
By the integral representation of $zetaleft(sright)$ we have $$zetaleft(2k+3right)-1=frac1left(2k+2right)!int_0^inftyfracu^2k+2e^uleft(e^u-1right)du$$ then $$sum_kgeq0left(zetaleft(2k+3right)-1right)=int_0^inftyfrac1e^uleft(e^u-1right)sum_kgeq0fracu^2k+2left(2k+2right)!du$$ $$=int_0^inftyfraccoshleft(uright)-1e^uleft(e^u-1right)du=frac12int_0^inftye^-u-e^-2udu=color redfrac14.$$
answered Mar 30 at 15:51
Marco Cantarini
28.8k23272
add a comment |
up vote
3
down vote
By the integral representation of $zetaleft(sright)$ we have $$zetaleft(2k+3right)-1=frac1left(2k+2right)!int_0^inftyfracu^2k+2e^uleft(e^u-1right)du$$ then $$sum_kgeq0left(zetaleft(2k+3right)-1right)=int_0^inftyfrac1e^uleft(e^u-1right)sum_kgeq0fracu^2k+2left(2k+2right)!du$$ $$=int_0^inftyfraccoshleft(uright)-1e^uleft(e^u-1right)du=frac12int_0^inftye^-u-e^-2udu=color redfrac14.$$
answered Mar 30 at 15:51
Marco Cantarini
28.8k23272
add a comment |
up vote
3
down vote
up vote
3
down vote
By the integral representation of $zetaleft(sright)$ we have $$zetaleft(2k+3right)-1=frac1left(2k+2right)!int_0^inftyfracu^2k+2e^uleft(e^u-1right)du$$ then $$sum_kgeq0left(zetaleft(2k+3right)-1right)=int_0^inftyfrac1e^uleft(e^u-1right)sum_kgeq0fracu^2k+2left(2k+2right)!du$$ $$=int_0^inftyfraccoshleft(uright)-1e^uleft(e^u-1right)du=frac12int_0^inftye^-u-e^-2udu=color redfrac14.$$
answered Mar 30 at 15:51
Marco Cantarini
28.8k23272
By the integral representation of $zetaleft(sright)$ we have $$zetaleft(2k+3right)-1=frac1left(2k+2right)!int_0^inftyfracu^2k+2e^uleft(e^u-1right)du$$ then $$sum_kgeq0left(zetaleft(2k+3right)-1right)=int_0^inftyfrac1e^uleft(e^u-1right)sum_kgeq0fracu^2k+2left(2k+2right)!du$$ $$=int_0^inftyfraccoshleft(uright)-1e^uleft(e^u-1right)du=frac12int_0^inftye^-u-e^-2udu=color redfrac14.$$
answered Mar 30 at 15:51
Marco Cantarini
28.8k23272
answered Mar 30 at 15:51
Marco Cantarini
28.8k23272
answered Mar 30 at 15:51
Marco Cantarini
28.8k23272
answered Mar 30 at 15:51
Marco Cantarini
28.8k23272
28.8k23272
add a comment |
add a comment |
up vote
0
down vote
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandroot[2],sqrt[#1],#2,,
newcommandtotald[3]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
&bbox[10px,#ffd]dssum_k = 0^inftybrackszetapars2k + 3 - 1 =
sum_k = 1^inftybrackszetapars2k + 1 - 1
\[5mm] = &
lim_z to 1bracks1 over 2z - 1 over 2,pi,cotparspi z -
1 over 1 - z^2 + 1 - gamma - Psipars1 + z = bbx1 over 4
endalign
where $dsPsi$ is the Digamma Function. See $dsmathbfcolorblack6.3.15$ in A & S Table.
Note that
$dsPsipars2 = Psipars1 + 1 = -gamma + 1$.
answered Apr 2 at 3:41
Felix Marin
65.6k7107138
add a comment |
up vote
0
down vote
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandroot[2],sqrt[#1],#2,,
newcommandtotald[3]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
&bbox[10px,#ffd]dssum_k = 0^inftybrackszetapars2k + 3 - 1 =
sum_k = 1^inftybrackszetapars2k + 1 - 1
\[5mm] = &
lim_z to 1bracks1 over 2z - 1 over 2,pi,cotparspi z -
1 over 1 - z^2 + 1 - gamma - Psipars1 + z = bbx1 over 4
endalign
where $dsPsi$ is the Digamma Function. See $dsmathbfcolorblack6.3.15$ in A & S Table.
Note that
$dsPsipars2 = Psipars1 + 1 = -gamma + 1$.
answered Apr 2 at 3:41
Felix Marin
65.6k7107138
add a comment |
up vote
0
down vote
up vote
0
down vote
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandroot[2],sqrt[#1],#2,,
newcommandtotald[3]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
&bbox[10px,#ffd]dssum_k = 0^inftybrackszetapars2k + 3 - 1 =
sum_k = 1^inftybrackszetapars2k + 1 - 1
\[5mm] = &
lim_z to 1bracks1 over 2z - 1 over 2,pi,cotparspi z -
1 over 1 - z^2 + 1 - gamma - Psipars1 + z = bbx1 over 4
endalign
where $dsPsi$ is the Digamma Function. See $dsmathbfcolorblack6.3.15$ in A & S Table.
Note that
$dsPsipars2 = Psipars1 + 1 = -gamma + 1$.
answered Apr 2 at 3:41
Felix Marin
65.6k7107138
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandroot[2],sqrt[#1],#2,,
newcommandtotald[3]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
&bbox[10px,#ffd]dssum_k = 0^inftybrackszetapars2k + 3 - 1 =
sum_k = 1^inftybrackszetapars2k + 1 - 1
\[5mm] = &
lim_z to 1bracks1 over 2z - 1 over 2,pi,cotparspi z -
1 over 1 - z^2 + 1 - gamma - Psipars1 + z = bbx1 over 4
endalign
where $dsPsi$ is the Digamma Function. See $dsmathbfcolorblack6.3.15$ in A & S Table.
Note that
$dsPsipars2 = Psipars1 + 1 = -gamma + 1$.
answered Apr 2 at 3:41
Felix Marin
65.6k7107138
edited Sep 11 at 3:04
answered Apr 2 at 3:41
Felix Marin
65.6k7107138
answered Apr 2 at 3:41
Felix Marin
65.6k7107138
answered Apr 2 at 3:41
Felix Marin
65.6k7107138
65.6k7107138
add a comment |
add a comment |
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2
Where did you found it? All terms are greater than 1!
– gammatester
Mar 30 at 13:41
2
You have a typo, should be $sum_k=0^infty (zeta(2k+3) - colorred1) = frac14$.
– achille hui
Mar 30 at 13:50
Actually , this sum seems to be $k+frac54$ with a very small error already for $k=10$
– Peter
Mar 30 at 13:51