Sunday, 25 August 2013

calculus - Evaluating series of zeta values like suminftyk=1fraczeta(2k)k16k=ln(pi)frac32ln(2)



Somehow I derived these values a few years ago but I forgot how.
It cannot be very hard (certainly doesn't require "advanced" knowledge) but I just don't know where to start.




Here are the sums:
k=1ζ(2k)4k=12k=1ζ(2k)16k=4π8k=1ζ(2k)k4k=ln(π)ln(2)k=1ζ(2k)k16k=ln(π)32ln(2).


Answer



Hint. One may start with the classic series expansion, which may come from the Weierstrass infinite product of the sine function,




n=1x2n2+x2=12(1+πxcot(πx)),|x|<1.





Expanding the left hand side of (1) one deduces




k=1ζ(2k)x2k=12(1πxcot(πx)),|x|<1.





By dividing (2) by x and integrating one gets




k=1ζ(2k)x2kk=log(πxsin(πx)),|x|<1.




Your equalities are now obtained by putting x:=12,14 in (2) and in (3).


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...