Vtyjkyuk

Trying to prove

$$zeta(2) - beta(1) + zeta(4) - beta(3) + zeta(6)- beta(5) + ldots=1$$
I found by numerical calculation that (when $k$ goes to infinity)

$$sum_n=1^kzeta (2n)=k+3/4+o(1),$$

$$sum_n=1^kbeta (2n-1)=(k-1)+3/4+o(1).$$

in order to prove the above formula. Now how can I prove it analytically?

We have:
$$zeta(2k) = frac1(2k-1)!int_0^+inftyfract^2k-1e^t-1 tag1 $$
and:
$$beta(2k-1)=sum_ngeq 0frac(-1)^n(2n+1)^2k-1=frac1(2k-2)!int_0^+inftyfract^2k-2e^te^2t+1,dt tag2$$
hence:
$$sum_ngeq 1 beta(2n-1), z^2n-1 = fracpi z4cdotsecfracpi z2,tag3 $$
$$sum_ngeq 1 zeta(2n),z^2n = frac1-pi z cot(pi z)2,tag4 $$
and:
$$ lim_zto 1^-sum_ngeq 1left(zeta(2n), z^2n-beta(2n-1), z^2n-1right) = colorred1.tag5$$

$underlineleft(1^textstright)colon$

$$
beginalign
S&=sum_n=1^inftyleft[zeta(2n)colorred-1+1-beta(2n-1)right] ,=colorMagentasum_n=1^inftyleft[zeta(2n)-1right],+,colorbluesum_n=1^inftyleft[1-beta(2n-1)right] \[4mm]
colorMagentaS_1&=sum_n=1^inftyleft[zeta(2n)-1right] ,=sum_n=1^inftysum_m=colorred2^inftyleft[frac1m^2nright] ,=sum_m=colorred2^inftyfrac1m^2-1 ,=sum_m=colorred2^inftyfrac1(m-1)(m+1) ,=colorMagentafrac34 \[4mm]
colorblueS_2&=sum_n=1^inftyleft[1-beta(2n-1)right] ,=sum_n=1^inftysum_m=colorred2^inftyleft[frac(-1)^m(2m-1)^2n-1right] ,=sum_m=colorred1^inftysum_n=1^inftyleft[smallfrac4m-1(4m-1)^2n-smallfrac4m+1(4m+1)^2nright] \
&=sum_m=1^inftyleft[smallfrac4m-1(4m-1)^2-1-smallfrac4m+1(4m+1)^2-1right] ,=sum_m=1^inftyfrac18m^2-2 ,=sum_m=1^inftyfracsmall 1/8(m-small 1/2)(m+small 1/2) ,=colorbluefrac14 \[4mm]
colorredS&=frac34,+,frac14 ,=colorred1
endalign
$$

$underlineleft(2^textndright)colon$

$$
beginalign
S&=sum_n=1^inftyleft[-beta(2n-1)+zeta(2n)right] ,=-beta(1)+sum_n=1^infty-1left[zeta(2n)-beta(2n+1)right]+lim_small nrightarrowinftyzeta(2n) \
&=colorgreenleft[lim_small nrightarrowinftyzeta(2n)-beta(1)right] +colorMagentasum_n=1^inftyleft[zeta(2n)-zeta(2n+1)right] +colorbluesum_n=1^inftyleft[zeta(2n+1)-beta(2n+1)right] \[6mm]
colorMagentaS_1&=sum_n=1^inftyleft[zeta(2n)-zeta(2n+1)right] ,=sum_n=1^inftysum_m=2^inftyleft(frac1m^2n-fracsmall 1/mm^2nright) \
&=sum_m=2^inftyleft(fracm-1mright)left(frac1m^2-1right) ,=sum_m=2^inftyfrac1m(m+1) ,=colorMagentafrac12 \[6mm]
colorblueS_2&=sum_n=1^inftyleft[zeta(2n+1)-beta(2n+1)right] ,=sum_n=1^inftyfracGamma(2n+1)zeta(2n+1),-,Gamma(2n+1)beta(2n+1)(2n)! \
&=int_0^inftyleft(frac1e^x-1-frace^xe^2x+1right)left(smallsum_n=1^inftyfracx^2n(2n)!right)dx =int_0^inftyleft(frac1e^x-1-frace^xe^2x+1right)left(smallcoshx-1right)dx \[2mm]
&=int_0^inftyleft(frace^x+1e^2x-1-frace^xe^2x+1right)left(frace^2x+12e^x-1right),dx ,=int_0^inftyleft(frac(e^x+1)^2e^4x-1,frac(e^x-1)^22e^xright),dx \[2mm]
&=int_0^inftyleft(frace^2x-1e^2x+1,fracsmall 1/2e^xright),dx ,=int_0^inftyleft(frace^xe^2x+1-fracsmall 1/2e^xright),dx ,=colorbluebeta(1)-frac12 \[6mm]
colorredS&=left[,lim_small nrightarrowinftyzeta(2n)-beta(1),right],+,left[,frac12,right],+,left[,beta(1)-frac12,right] ,=lim_small nrightarrowinftyzeta(2n) ,=colorred1
endalign
$$

$underlineleft(textNBright)colon$

$$
beginalign
colorredS&=sum_n=1^inftyleft[zeta(2n)-beta(2n-1)right] =sum_n=1^inftyleft[fracGamma(2n),zeta(2n)(2n-1)!-fracGamma(2n-1),beta(2n-1)(2n-2)!right] \[2mm]
&=sum_n=1^inftyint_0^inftyleft[fracx^2n-1(2n-1)!frac1e^x-1,-,fracx^2n-2(2n-2)!frace^xe^2x+1right],dx \[2mm]
&=int_0^inftyleft[frac1e^x-1left(sum_n=1^inftyfracx^2n-1(2n-1)!right),-,frace^xe^2x+1left(sum_n=1^inftyfracx^2n-2(2n-2)!right)right],dx \[2mm]
&=int_0^inftyleft[fracsinh(x)e^x-1,-,frace^xcosh(x)e^2x+1right],dx =frac12int_0^inftyfracdxe^x =colorredfrac12,rightarrow,mathcalX,,smalltextincorrect!
endalign
$$