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=12∞∑k=1ζ(2k)16k=4−π8∞∑k=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(1−πxcot(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).
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