Vtyjkyuk

I am aware that I can solve the $$sum_n=1^inftyfrac1n^4,$$ using a a cosine series for $x^2$ on the half period $0<x<2$ however I am wondering if I can also solve this by using the cosine half series for x and applying parsavels identity.

The result gives a series of $$ int_0^2x^2 dx= b+asum_n=0^inftyfrac1(2n+1)^4,$$ where b and a are real numbers, thus giving me a sum for the odd terms.
I am unsure how to compute the even terms with this method any help would be appreciated

It is worth noting that, generalization to any power (i.e., other than the given case 4!) is possible here.
The Riemann Zeta function $zeta(s)$ and the complementary function Dirichlet alternating zeta function $eta(s)$

begineqnarray*
zeta(s) &=& sum_n=1^inftyfrac1n^s \
eta(s) &=& -sum_n=1^inftyfrac(-1)^nn^s
endeqnarray*

are related by
begineqnarray
eta(s) &=& left(1-2^1-sright)zeta(s)
endeqnarray,

Also,

begineqnarray*
2 sum_m=1^inftyfrac1(2 m)^s &=& sum_n=1^inftyfrac1n^s + sum_n=1^inftyfrac(-1)^nn^s \
&=& 2 left(2^-sright) zeta(s) \
2 sum_m=1^inftyfrac1(2 m-1)^s &=& sum_n=1^inftyfrac1n^s - sum_n=1^inftyfrac(-1)^nn^s \
&=& 2 left(1-2^-sright) zeta(s)
endeqnarray*

The odd and even sum respectively becomes,

begineqnarray*
sum_m=1^inftyfrac1(2 m)^s &=& 2^-szeta(s) \
sum_m=1^inftyfrac1(2 m-1)^s &=& left(1-2^-sright) zeta(s)
endeqnarray*

In the particular case $s=4$, it becomes $frac116 zeta(4)$ and $frac1516 zeta(4)$ as Jean-Claude Arbaut arrived at.