The Kronecker Symbol $\left(\frac nm\right)$ has a range of $\{-1,0,1\}$ and $\sum\limits_{n=1}^\infty\frac{(-1)^n}n=-\ln(2)$, so we combine to find the following with the using software. Also note the DirichletL function $\text L_{k,j}(s)$ from the link:
$$\sum_{n=1}^\infty \frac {\left(\frac 2n\right)}n=\sum_{n=1}^\infty \frac {\left(\frac n2\right)}n=\text L_{8,2}(1)=\frac{\sinh^{-1}(1)}{\sqrt2}$$
which seems true
$$\sum_{n=1}^\infty \frac {\left(\frac n3\right)}n=\text L_{3,2}(1)=\frac\pi{3\sqrt 3}$$
which also seems true
$$\sum_{n=1}^\infty \frac {\left(\frac n4\right)}n, \sum_{n=1}^\infty \frac {\left(\frac 4n\right)}n\text{ diverge}$$
$$\sum_{n=1}^\infty \frac {\left(\frac n5\right)} n=\frac{\ln(\varphi+1)}{\sqrt5}$$
which is probably correct. Note the golden ratio
$$\sum_{n=1}^\infty \frac {\left(\frac n6\right)} n\mathop=^?\frac\pi{\sqrt 6} $$
which may be true. Note the golden ratio. Are there any closed forms of $\sum\limits_{n=1}^\infty \frac {\left(\frac n x\right)}n$ and $\sum\limits_{n=1}^\infty \frac {\left(\frac xn\right)}n$?
Best Answer
These are values at $s = 1$ of the $L$ functions $L(\chi, s)$ attached to the quadratic characters $\chi(n) = \left(\frac m n\right)$.
Depending on the parity of $\chi$ (i.e. whether $\chi(-1)$ is equal to $1$ or $-1$, where we view $\chi$ as a Dirichlet character), they are either a product of $\pi$ with simple factors (evaluated as special values of Hurwitz zeta functions) or something involving the $\log$ of a fundamental unit (the regulator) as given by class number formula.
I will not copy all the details here, but the above wiki pages should be good enough as starting points. Also these results should be available in many books on algebraic/analytic number theory, although they might not be listed together as the techniques are different for the two parities.