How to show that $gamma=limlimits_n to inftyleft(sumlimits_k=1^nzeta(2k)over kright)-ln(2n)$
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Proposed:
$$gamma=lim_n to inftyleftleft[,sum_k=1^nzetaleft(,2k,right)over k,right] -lnleft(,2n,right)righttag1$$
Where $gamma$ is Euler-Mascheroni Constant
My try
Well-known $$gamma=lim_nto inftyleft(sum_k=1^n1over kright)-ln(n)tag2$$
How do we prove $(1)?$
sequences-and-series limits zeta-functions eulers-constant
edited May 13 '17 at 3:43
Felix Marin
65.6k7107138
asked May 8 '17 at 12:04
gymbvghjkgkjkhgfkl
1
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up vote
4
down vote
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Proposed:
$$gamma=lim_n to inftyleftleft[,sum_k=1^nzetaleft(,2k,right)over k,right] -lnleft(,2n,right)righttag1$$
Where $gamma$ is Euler-Mascheroni Constant
My try
Well-known $$gamma=lim_nto inftyleft(sum_k=1^n1over kright)-ln(n)tag2$$
How do we prove $(1)?$
sequences-and-series limits zeta-functions eulers-constant
edited May 13 '17 at 3:43
Felix Marin
65.6k7107138
asked May 8 '17 at 12:04
gymbvghjkgkjkhgfkl
1
Anyway for the question use $$pi;x;cot(pi;x)-1=-2sum_k=1^infty zeta(2k);x^2k$$
– Zaid Alyafeai
May 8 '17 at 12:15
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Proposed:
$$gamma=lim_n to inftyleftleft[,sum_k=1^nzetaleft(,2k,right)over k,right] -lnleft(,2n,right)righttag1$$
Where $gamma$ is Euler-Mascheroni Constant
My try
Well-known $$gamma=lim_nto inftyleft(sum_k=1^n1over kright)-ln(n)tag2$$
How do we prove $(1)?$
sequences-and-series limits zeta-functions eulers-constant
edited May 13 '17 at 3:43
Felix Marin
65.6k7107138
asked May 8 '17 at 12:04
gymbvghjkgkjkhgfkl
1
Proposed:
$$gamma=lim_n to inftyleftleft[,sum_k=1^nzetaleft(,2k,right)over k,right] -lnleft(,2n,right)righttag1$$
Where $gamma$ is Euler-Mascheroni Constant
My try
Well-known $$gamma=lim_nto inftyleft(sum_k=1^n1over kright)-ln(n)tag2$$
How do we prove $(1)?$
sequences-and-series limits zeta-functions eulers-constant
sequences-and-series limits zeta-functions eulers-constant
edited May 13 '17 at 3:43
Felix Marin
65.6k7107138
asked May 8 '17 at 12:04
gymbvghjkgkjkhgfkl
1
edited May 13 '17 at 3:43
Felix Marin
65.6k7107138
asked May 8 '17 at 12:04
gymbvghjkgkjkhgfkl
1
edited May 13 '17 at 3:43
Felix Marin
65.6k7107138
edited May 13 '17 at 3:43
Felix Marin
65.6k7107138
edited May 13 '17 at 3:43
Felix Marin
65.6k7107138
65.6k7107138
asked May 8 '17 at 12:04
gymbvghjkgkjkhgfkl
1
asked May 8 '17 at 12:04
gymbvghjkgkjkhgfkl
1
asked May 8 '17 at 12:04
gymbvghjkgkjkhgfkl
1
1
Anyway for the question use $$pi;x;cot(pi;x)-1=-2sum_k=1^infty zeta(2k);x^2k$$
– Zaid Alyafeai
May 8 '17 at 12:15
add a comment |
Anyway for the question use $$pi;x;cot(pi;x)-1=-2sum_k=1^infty zeta(2k);x^2k$$
– Zaid Alyafeai
May 8 '17 at 12:15
Anyway for the question use $$pi;x;cot(pi;x)-1=-2sum_k=1^infty zeta(2k);x^2k$$
– Zaid Alyafeai
May 8 '17 at 12:15
Anyway for the question use $$pi;x;cot(pi;x)-1=-2sum_k=1^infty zeta(2k);x^2k$$
– Zaid Alyafeai
May 8 '17 at 12:15
add a comment |
2 Answers
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Note that $$sum_kgeq1fraczetaleft(2kright)-1k=sum_kgeq1frac1ksum_mgeq2frac1m^2k$$ $$=sum_mgeq2sum_kgeq1frac1km^2k=-sum_mgeq2logleft(1-frac1m^2right)$$ $$=-logleft(prod_mgeq2left(1-frac1m^2right)right)$$ and it is well known that $$prod_mgeq2left(1-frac1m^2right)=frac12$$ since $$prod_m=2^Nleft(1-frac1m^2right)=prod_m=2^Nfracm^2-1m^2=prod_m=2^Nfracm-1mprod_m=2^Nfracm+1m=frac1NfracN+12$$ so $$sum_kgeq1fraczetaleft(2kright)-1k=colorredlogleft(2right).$$ The result obviously implies your limit since $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-logleft(2nright)right)=lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-sum_k=1^nfrac1k+sum_k=1^nfrac1k-logleft(2nright)right)$$ $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)-1k+sum_k=1^nfrac1k-logleft(nright)-logleft(2right)right)=lim_nrightarrowinftyleft(sum_k=1^nfrac1k-logleft(nright)right)=colorbluegamma.$$
answered May 8 '17 at 16:44
Marco Cantarini
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Using the given formula (2) it is sufficient to prove that $$sum_k=1^infty frac1k (zeta(2k) - 1 ) = ln 2.$$ So now set $f(x) := sum_k=1 ^infty frac1k (zeta(2k) - 1 ) x^2k$. Then $$xf'(x) = 2 sum_k=1^infty (zeta(2k) - 1 ) x^2k = - pi x cot(pi x ) + 1 - frac2x^21 - x^2.$$ Therefore, integrating $f'(x)$, $$f(1-epsilon) - f(epsilon) = ln(1-epsilon) - ln(1 -epsilon^2) +ln(2 - epsilon)$$ for $0<epsilon<1$. Letting $epsilon to 0$ we get the desired conclusion.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Note that $$sum_kgeq1fraczetaleft(2kright)-1k=sum_kgeq1frac1ksum_mgeq2frac1m^2k$$ $$=sum_mgeq2sum_kgeq1frac1km^2k=-sum_mgeq2logleft(1-frac1m^2right)$$ $$=-logleft(prod_mgeq2left(1-frac1m^2right)right)$$ and it is well known that $$prod_mgeq2left(1-frac1m^2right)=frac12$$ since $$prod_m=2^Nleft(1-frac1m^2right)=prod_m=2^Nfracm^2-1m^2=prod_m=2^Nfracm-1mprod_m=2^Nfracm+1m=frac1NfracN+12$$ so $$sum_kgeq1fraczetaleft(2kright)-1k=colorredlogleft(2right).$$ The result obviously implies your limit since $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-logleft(2nright)right)=lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-sum_k=1^nfrac1k+sum_k=1^nfrac1k-logleft(2nright)right)$$ $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)-1k+sum_k=1^nfrac1k-logleft(nright)-logleft(2right)right)=lim_nrightarrowinftyleft(sum_k=1^nfrac1k-logleft(nright)right)=colorbluegamma.$$
answered May 8 '17 at 16:44
Marco Cantarini
28.8k23272
add a comment |
up vote
4
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Note that $$sum_kgeq1fraczetaleft(2kright)-1k=sum_kgeq1frac1ksum_mgeq2frac1m^2k$$ $$=sum_mgeq2sum_kgeq1frac1km^2k=-sum_mgeq2logleft(1-frac1m^2right)$$ $$=-logleft(prod_mgeq2left(1-frac1m^2right)right)$$ and it is well known that $$prod_mgeq2left(1-frac1m^2right)=frac12$$ since $$prod_m=2^Nleft(1-frac1m^2right)=prod_m=2^Nfracm^2-1m^2=prod_m=2^Nfracm-1mprod_m=2^Nfracm+1m=frac1NfracN+12$$ so $$sum_kgeq1fraczetaleft(2kright)-1k=colorredlogleft(2right).$$ The result obviously implies your limit since $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-logleft(2nright)right)=lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-sum_k=1^nfrac1k+sum_k=1^nfrac1k-logleft(2nright)right)$$ $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)-1k+sum_k=1^nfrac1k-logleft(nright)-logleft(2right)right)=lim_nrightarrowinftyleft(sum_k=1^nfrac1k-logleft(nright)right)=colorbluegamma.$$
answered May 8 '17 at 16:44
Marco Cantarini
28.8k23272
add a comment |
up vote
4
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up vote
4
down vote
accepted
Note that $$sum_kgeq1fraczetaleft(2kright)-1k=sum_kgeq1frac1ksum_mgeq2frac1m^2k$$ $$=sum_mgeq2sum_kgeq1frac1km^2k=-sum_mgeq2logleft(1-frac1m^2right)$$ $$=-logleft(prod_mgeq2left(1-frac1m^2right)right)$$ and it is well known that $$prod_mgeq2left(1-frac1m^2right)=frac12$$ since $$prod_m=2^Nleft(1-frac1m^2right)=prod_m=2^Nfracm^2-1m^2=prod_m=2^Nfracm-1mprod_m=2^Nfracm+1m=frac1NfracN+12$$ so $$sum_kgeq1fraczetaleft(2kright)-1k=colorredlogleft(2right).$$ The result obviously implies your limit since $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-logleft(2nright)right)=lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-sum_k=1^nfrac1k+sum_k=1^nfrac1k-logleft(2nright)right)$$ $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)-1k+sum_k=1^nfrac1k-logleft(nright)-logleft(2right)right)=lim_nrightarrowinftyleft(sum_k=1^nfrac1k-logleft(nright)right)=colorbluegamma.$$
answered May 8 '17 at 16:44
Marco Cantarini
28.8k23272
Note that $$sum_kgeq1fraczetaleft(2kright)-1k=sum_kgeq1frac1ksum_mgeq2frac1m^2k$$ $$=sum_mgeq2sum_kgeq1frac1km^2k=-sum_mgeq2logleft(1-frac1m^2right)$$ $$=-logleft(prod_mgeq2left(1-frac1m^2right)right)$$ and it is well known that $$prod_mgeq2left(1-frac1m^2right)=frac12$$ since $$prod_m=2^Nleft(1-frac1m^2right)=prod_m=2^Nfracm^2-1m^2=prod_m=2^Nfracm-1mprod_m=2^Nfracm+1m=frac1NfracN+12$$ so $$sum_kgeq1fraczetaleft(2kright)-1k=colorredlogleft(2right).$$ The result obviously implies your limit since $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-logleft(2nright)right)=lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-sum_k=1^nfrac1k+sum_k=1^nfrac1k-logleft(2nright)right)$$ $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)-1k+sum_k=1^nfrac1k-logleft(nright)-logleft(2right)right)=lim_nrightarrowinftyleft(sum_k=1^nfrac1k-logleft(nright)right)=colorbluegamma.$$
answered May 8 '17 at 16:44
Marco Cantarini
28.8k23272
edited May 8 '17 at 16:50
answered May 8 '17 at 16:44
Marco Cantarini
28.8k23272
answered May 8 '17 at 16:44
Marco Cantarini
28.8k23272
answered May 8 '17 at 16:44
Marco Cantarini
28.8k23272
28.8k23272
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add a comment |
up vote
3
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Using the given formula (2) it is sufficient to prove that $$sum_k=1^infty frac1k (zeta(2k) - 1 ) = ln 2.$$ So now set $f(x) := sum_k=1 ^infty frac1k (zeta(2k) - 1 ) x^2k$. Then $$xf'(x) = 2 sum_k=1^infty (zeta(2k) - 1 ) x^2k = - pi x cot(pi x ) + 1 - frac2x^21 - x^2.$$ Therefore, integrating $f'(x)$, $$f(1-epsilon) - f(epsilon) = ln(1-epsilon) - ln(1 -epsilon^2) +ln(2 - epsilon)$$ for $0<epsilon<1$. Letting $epsilon to 0$ we get the desired conclusion.
add a comment |
up vote
3
down vote
Using the given formula (2) it is sufficient to prove that $$sum_k=1^infty frac1k (zeta(2k) - 1 ) = ln 2.$$ So now set $f(x) := sum_k=1 ^infty frac1k (zeta(2k) - 1 ) x^2k$. Then $$xf'(x) = 2 sum_k=1^infty (zeta(2k) - 1 ) x^2k = - pi x cot(pi x ) + 1 - frac2x^21 - x^2.$$ Therefore, integrating $f'(x)$, $$f(1-epsilon) - f(epsilon) = ln(1-epsilon) - ln(1 -epsilon^2) +ln(2 - epsilon)$$ for $0<epsilon<1$. Letting $epsilon to 0$ we get the desired conclusion.
add a comment |
up vote
3
down vote
up vote
3
down vote
Using the given formula (2) it is sufficient to prove that $$sum_k=1^infty frac1k (zeta(2k) - 1 ) = ln 2.$$ So now set $f(x) := sum_k=1 ^infty frac1k (zeta(2k) - 1 ) x^2k$. Then $$xf'(x) = 2 sum_k=1^infty (zeta(2k) - 1 ) x^2k = - pi x cot(pi x ) + 1 - frac2x^21 - x^2.$$ Therefore, integrating $f'(x)$, $$f(1-epsilon) - f(epsilon) = ln(1-epsilon) - ln(1 -epsilon^2) +ln(2 - epsilon)$$ for $0<epsilon<1$. Letting $epsilon to 0$ we get the desired conclusion.
Using the given formula (2) it is sufficient to prove that $$sum_k=1^infty frac1k (zeta(2k) - 1 ) = ln 2.$$ So now set $f(x) := sum_k=1 ^infty frac1k (zeta(2k) - 1 ) x^2k$. Then $$xf'(x) = 2 sum_k=1^infty (zeta(2k) - 1 ) x^2k = - pi x cot(pi x ) + 1 - frac2x^21 - x^2.$$ Therefore, integrating $f'(x)$, $$f(1-epsilon) - f(epsilon) = ln(1-epsilon) - ln(1 -epsilon^2) +ln(2 - epsilon)$$ for $0<epsilon<1$. Letting $epsilon to 0$ we get the desired conclusion.
edited May 8 '17 at 14:18
answered May 8 '17 at 13:32
user438618
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Anyway for the question use $$pi;x;cot(pi;x)-1=-2sum_k=1^infty zeta(2k);x^2k$$
– Zaid Alyafeai
May 8 '17 at 12:15