How to show that $gamma=limlimits_n to inftyleft(sumlimits_k=1^nzeta(2k)over kright)-ln(2n)$
Clash Royale CLAN TAG#URR8PPP
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
add a comment |
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
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
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
65.6k7107138
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
2
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 $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-logleft(2nright)right)=lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-sum_k=1^nfrac1k+sum_k=1^nfrac1k-logleft(2nright)right)$$ $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)-1k+sum_k=1^nfrac1k-logleft(nright)-logleft(2right)right)=lim_nrightarrowinftyleft(sum_k=1^nfrac1k-logleft(nright)right)=colorbluegamma.$$
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 |
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 $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-logleft(2nright)right)=lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-sum_k=1^nfrac1k+sum_k=1^nfrac1k-logleft(2nright)right)$$ $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)-1k+sum_k=1^nfrac1k-logleft(nright)-logleft(2right)right)=lim_nrightarrowinftyleft(sum_k=1^nfrac1k-logleft(nright)right)=colorbluegamma.$$
add a comment |
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 $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-logleft(2nright)right)=lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-sum_k=1^nfrac1k+sum_k=1^nfrac1k-logleft(2nright)right)$$ $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)-1k+sum_k=1^nfrac1k-logleft(nright)-logleft(2right)right)=lim_nrightarrowinftyleft(sum_k=1^nfrac1k-logleft(nright)right)=colorbluegamma.$$
add a comment |
up vote
4
down vote
accepted
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 $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-logleft(2nright)right)=lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-sum_k=1^nfrac1k+sum_k=1^nfrac1k-logleft(2nright)right)$$ $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)-1k+sum_k=1^nfrac1k-logleft(nright)-logleft(2right)right)=lim_nrightarrowinftyleft(sum_k=1^nfrac1k-logleft(nright)right)=colorbluegamma.$$
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 $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-logleft(2nright)right)=lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-sum_k=1^nfrac1k+sum_k=1^nfrac1k-logleft(2nright)right)$$ $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)-1k+sum_k=1^nfrac1k-logleft(nright)-logleft(2right)right)=lim_nrightarrowinftyleft(sum_k=1^nfrac1k-logleft(nright)right)=colorbluegamma.$$
edited May 8 '17 at 16:50
answered May 8 '17 at 16:44
Marco Cantarini
28.8k23272
28.8k23272
add a comment |
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
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
add a comment |
add a comment |
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2271380%2fhow-to-show-that-gamma-lim-limits-n-to-infty-left-sum-limits-k-1n%23new-answer', 'question_page');
);
Post as a guest
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
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
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
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