Thursday, 26 April 2018

sequences and series - Proving that fracpi332=1suminftyk=1frac2k(2k+1)zeta(2k+2)42k+2




After numerical analysis it seems that



π332=1k=12k(2k+1)ζ(2k+2)42k+2



Could someone prove the validity of such identity?


Answer



Yes, we can prove it. We can change the order of summation in




k=12k(2k+1)ζ(2k+2)42k+2=k=12k(2k+1)42k+2n=11n2k+2=n=1k=12k(2k+1)(4n)2k+2=n=1r(4n),



where, for |z|>1, we define




r(z)=k=11z2k=1z21=12(1z11z+1).



Differentiating yields r(z)=1(z1)31(z+1)3, so



1k=12k(2k+1)ζ(2k+2)42k+2=ν=0(1)ν(2ν+1)3,



and the latter sum is by an earlier answer using the partial fraction decomposition of 1cosz:



ν=0(1)ν(2ν+1)3=π332E2=π332.


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