For each natural number $a$ consider the sequence $l(a):=\left(\frac{\gcd(a,b)}{a+b}\right)\_{b \in \mathbb{N}}$.


Then I have computed that for $k\ge 2, k \in \mathbb{R}$ and $p$ prime, we have:


$$|l(1)|\_k^k = \zeta(k)-1$$


$$|l(p)|\_k^k = \frac{2 p^k-1}{p^k}\zeta(k)-\left(1+\sum\_{j=1}^{p-1} \frac{1}{j^k}\right)$$


I have also computed for $n=4$ that:


$$|l(4)|\_k^k = \zeta(k)\left(3-\frac{1}{4^k}-\frac{2}{2^k}+\frac{1}{2^{2k}}\right)-3-\frac{1}{3^k}$$


My question is, if $|l(a)|\_k^k = x \zeta(k) + y$ where $x,y \in \mathbb{Q}$?


Furthermore:


$$\langle l(1),l(2)\rangle = \sum\_{k=1}^\infty \frac{3k+1}{2k(k+1)(2k+1)}$$


Is this last quantity equal to $\log(2)$?



Can this quantity be expressed as $x\cdot \zeta(k)+y, x,y \in \mathbb{Q}$?
Yes, and yes. 


1) We have


$$|l(a)|\_k^k=\sum\_{d\mid a}\left(\prod\_{p\mid d}\left(1-\dfrac{1}{p^k}\right)\right)\left(\zeta(k)-\sum\_{j=1}^{a/d}\dfrac{1}{j^k}\right)\;,$$
which checks with your special cases.


2) Expand into partial fractions.


P.S. The coefficient of $\zeta(k)$ is equal to
$$\prod\_{p^v\Vert a}\left(v+1-\dfrac{v}{p^k}\right)\;,$$
for instance $3-2/2^k$ for $a=4$.


