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 $(1)?$










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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















up 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 $(1)?$










share|cite|improve this question























  • 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













up vote
4
down vote

favorite
3









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 $(1)?$










share|cite|improve this question















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






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edited May 13 '17 at 3:43









Felix Marin

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asked May 8 '17 at 12:04









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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

















  • 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











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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 $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-logleft(2nright)right)=lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-sum_k=1^nfrac1k+sum_k=1^nfrac1k-logleft(2nright)right)$$ $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)-1k+sum_k=1^nfrac1k-logleft(nright)-logleft(2right)right)=lim_nrightarrowinftyleft(sum_k=1^nfrac1k-logleft(nright)right)=colorbluegamma.$$






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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
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      2 Answers
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      active

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      active

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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 $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-logleft(2nright)right)=lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-sum_k=1^nfrac1k+sum_k=1^nfrac1k-logleft(2nright)right)$$ $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)-1k+sum_k=1^nfrac1k-logleft(nright)-logleft(2right)right)=lim_nrightarrowinftyleft(sum_k=1^nfrac1k-logleft(nright)right)=colorbluegamma.$$






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        up vote
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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 $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-logleft(2nright)right)=lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-sum_k=1^nfrac1k+sum_k=1^nfrac1k-logleft(2nright)right)$$ $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)-1k+sum_k=1^nfrac1k-logleft(nright)-logleft(2right)right)=lim_nrightarrowinftyleft(sum_k=1^nfrac1k-logleft(nright)right)=colorbluegamma.$$






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          up vote
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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 $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-logleft(2nright)right)=lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-sum_k=1^nfrac1k+sum_k=1^nfrac1k-logleft(2nright)right)$$ $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)-1k+sum_k=1^nfrac1k-logleft(nright)-logleft(2right)right)=lim_nrightarrowinftyleft(sum_k=1^nfrac1k-logleft(nright)right)=colorbluegamma.$$






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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 $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-logleft(2nright)right)=lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)k-sum_k=1^nfrac1k+sum_k=1^nfrac1k-logleft(2nright)right)$$ $$lim_nrightarrowinftyleft(sum_k=1^nfraczetaleft(2kright)-1k+sum_k=1^nfrac1k-logleft(nright)-logleft(2right)right)=lim_nrightarrowinftyleft(sum_k=1^nfrac1k-logleft(nright)right)=colorbluegamma.$$







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          edited May 8 '17 at 16:50

























          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.






              share|cite|improve this answer


























                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.






                share|cite|improve this answer
























                  up vote
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                  up vote
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                  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.






                  share|cite|improve this answer














                  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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                  edited May 8 '17 at 14:18

























                  answered May 8 '17 at 13:32







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