# Evaluate a kronecker symbol sum: $\sum\limits_{n=1}^\infty \frac {\big(\frac n x\big)}n$ and $\sum\limits_{n=1}^\infty \frac {\big(\frac xn\big)}n$

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 $$\sum\limits_{n=1}^\infty \frac {\left(\frac xn\right)}n$$?

## 1 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.

• I looked at their documentation: on this page they claim that "DirichletCharacter[k,j,n] picks a particular ordering for possible Dirichlet characters modulo k". I don't find anywhere what the "particular ordering" is, thus am not able to tell what $j$ is - you might have to do some test. The modulus $k$ should be $m$ or $4m$. Jun 26 at 1:13
• There is also a book Primes of the form $x^2 + ny^2$ by David Cox, in which he talks about this kind of characters around Lemma 1.14. Jun 26 at 1:14
• @TymaGaidash $L_{k,j}(s)=\sum\limits_{n=1}^\infty \frac{\chi_{k,j}(n)}{n^s}$ so evaluating at $s=1$ leads to $L_{k,j}(1)=\sum\limits_{n=1}^\infty \frac{\chi_{k,j}(n)}{n}$. Now for example $\chi_{5,3}(n)=\left(\frac{5}{n}\right)=\{1,-1,-1,1,0,...\}$ which repeats with period $5$, so $L_{5,3}(1)=\sum\limits_{n=1}^\infty \frac{\chi_{5,3}(n)}{n}=\sum\limits_{n=1}^\infty \frac{\left(\frac{5}{n}\right)}{n}=\frac{1}{5} \left(\sqrt{5} \log \left(\frac{\sqrt{5}}{8}+\frac{5}{8}\right)-\sqrt{5} \log \left(\frac{5}{8}-\frac{\sqrt{5}}{8}\right)\right)$. Jul 3 at 23:55
• @TymaGaidash Also $\left(\frac{5}{n}\right)=\left(\frac{n}{5}\right)$ so $L_{5,3}(1)=\sum\limits_{n=1}^\infty \frac{\left(\frac{n}{5}\right)}{n}=\frac{1}{5} \left(\sqrt{5} \log \left(\frac{\sqrt{5}}{8}+\frac{5}{8}\right)-\sqrt{5} \log \left(\frac{5}{8}-\frac{\sqrt{5}}{8}\right)\right)\approx 0.430409$. Jul 4 at 0:17
• @TymaGaidash You can explore Dirichlet characters on Wolfram alpha (see wolframalpha.com/… ) and also LMFDB is a good resource (see lmfdb.org/Character/Dirichlet/…) but the two sites may use different indexing schemes with respect to $j$. Jul 4 at 4:29