# A conjectural infinite series for $\frac{\pi^2}{5\sqrt{5}}$

I am looking for a proof of the following claim:

First define the function $$\chi(n)$$ as follows: $$\chi(n)=\begin{cases}1, & \text{if }n \equiv \pm 1 \pmod{10} \\ -1, & \text{if }n \equiv \pm 3 \pmod{10} \\ 0, & \text{if otherwise } \end{cases}$$

Then, $$\frac{\pi^2}{5\sqrt{5}}=\displaystyle\sum_{n=1}^{\infty}\frac{\chi(n)}{n^2}$$

The SageMath cell that demonstrates this claim can be found here.

More generally, if $$1\le k\le N-1$$ is an integer, where $$N$$ is a positive interger, $$S_{N,k} := \sum_{n=0}^\infty\biggl( \frac{1}{(N n+N-k)^2} + \frac{1}{(N n+k)^2} \biggr) = \frac{\pi^2}{N^2\sin^2(\pi k/N)}.$$ Your sum is $$S_{10,1}-S_{10,3}$$.

• needs reference Commented Sep 2, 2021 at 15:15
• Evaluate digamma of a rational number; Gauss's digamma theorem en.wikipedia.org/wiki/… Commented Sep 2, 2021 at 18:31

This is asking for the value of an $$L$$-function of an even Dirichlet character $$\chi$$ at a positive even integer, and these have known values. It is analogous to the explicit expressions for the Riemann zeta-function at positive even integers (definitely not at positive odd integers!), but instead of the values being rational multiples of powers of $$\pi$$, they are algebraic multiples of powers of $$\pi$$.

The particular Dirichlet character $$\chi$$ in this question is defined modulo 10, but it is a lifting of a character mod 5. Let $$\psi(n) = (\frac{n}{5})$$, which is a nontrivial even Dirichlet character mod 5. For odd $$n$$, $$\chi(n) = \psi(n)$$, so for $${\rm Re}(s) > 1$$, $$L(s,\chi) = \sum_{n \geq 1} \frac{\chi(n)}{n^s} = \prod_{p > 2} \frac{1}{1 - \psi(p)/p^s} = \left(1 - \frac{\psi(2)}{2^s}\right)L(s,\psi) = \left(1 + \frac{1}{2^s}\right)L(s,\psi).$$ Taking $$s = 2$$, $$\sum_{n \geq 1} \frac{\chi(n)}{n^2} = \frac{5}{4}L(2,\psi).$$ I will show $$L(2,\psi) = \frac{4}{25\sqrt{5}}\pi^2$$, and multiplying this by $$5/4$$ gives the desired value for $$\sum \chi(n)/n^2$$.

The character $$\psi \bmod 5$$ is primitive since every nontrivial Dirichlet character modulo a prime is primitive. It is a quadratic character: its nonzero values are $$\pm 1$$. For each even primitive quadratic Dirichlet character $$\eta \bmod m$$, let's work out a formula for $$L(k,\eta)$$ when $$k$$ is a positive even integer and then apply it to $$\eta = \psi$$ and $$k = 2$$. (There are analogous formulas for $$L(k,\eta)$$ when $$\eta \bmod m$$ is an odd primitive quadratic Dirichlet character and $$k$$ is a positive odd integer, and also formulas when $$\eta$$ is not quadratic, but I omit all of this for simplicity.)

For even primitive quadratic $$\eta \bmod m$$ and $${\rm Re}(s) > 1$$, the completed $$L$$-function of $$\eta$$ is $$\Lambda(s,\eta) := \left(\frac{\pi}{m}\right)^{-s/2}\Gamma\left(\frac{s}{2}\right)L(s,\eta).$$ This turns out to be an entire function, so $$L(s,\eta)$$ is also entire, and we have a functional equation $$\Lambda(s,\eta) = \Lambda(1-s,\eta)$$. (If $$\eta$$ were not quadratic, the functional equation would be more complicated.) When $$s = k$$ is a positive even integer, unraveling the formula $$\Lambda(k,\eta) = \Lambda(1-k,\eta)$$ gives us $$\left(\frac{\pi}{m}\right)^{-k/2}\Gamma\left(\frac{k}{2}\right)L(k,\eta) = \left(\frac{\pi}{m}\right)^{-(1-k)/2}\Gamma\left(\frac{1-k}{2}\right)L(1-k,\eta),$$ so $$L(k,\eta) = \left(\frac{\pi}{m}\right)^{k - 1/2}\frac{\Gamma((1-k)/2)}{\Gamma(k/2)}L(1-k,\eta).$$ The first factor on the right is a known expression. Using the reflection and duplication formulas for the Gamma-function, $$\frac{\Gamma((1-s)/2))}{\Gamma(s/2)} = \frac{2^{s-1}\sqrt{\pi}}{\cos\left(\frac{\pi}{2}s\right)\Gamma(s)}.$$ Letting $$s = k$$ be an even integer, $$\cos((\pi/2)k) = \cos((k/2)\pi) = (-1)^{k/2}$$, so $$L(k,\eta) = \left(\frac{\pi}{m}\right)^{k - 1/2}\frac{(-1)^{k/2}2^{k-1}\sqrt{\pi}}{(k-1)!}L(1-k,\eta) = \frac{(-1)^{k/2}2^{k-1}}{(k-1)!m^{k-1/2}}L(1-k,\eta)\pi^k$$ and the number $$L(1-k,\eta)$$ for positive integers $$k$$ and nontrivial primitive $$\eta \bmod m$$ turns out to be algebraic. (That generalizes the Riemann zeta-function at negative integers being rational). Explicitly, $$L(1-k,\eta) = -\frac{B_{k,\eta}}{k}$$ where the "twisted" Bernoulli number $$B_{k,\eta}$$ has exponential generating function $$\sum_{k \geq 0} B_{k,\eta}\frac{x^k}{k!} = \sum_{j=1}^m \eta(j)\frac{xe^{jx}}{e^{mx}-1},$$ so $$B_{k,\eta}$$ is a $$\mathbf Q$$-linear combination of values of $$\eta$$. (See Larry Washington's "Introduction to Cyclotomic Fields", esp. Theorem 4.2.)

Let's specialize all this to the case when $$\eta$$ is the (even primitive) quadratic character $$\psi \bmod 5$$ above and $$k = 2$$. We get $$L(2,\psi) = \frac{-2}{5^{3/2}}L(-1,\eta)\pi^2 = \frac{B_{2,\psi}}{5\sqrt{5}}\pi^2.$$ We had said the coefficient of $$\pi^2$$ turns out to be $$4/(25\sqrt{5})$$, and we'll get this by showing $$B_{2,\psi} = 4/5$$.

For the character $$\psi \bmod 5$$, $$\sum_{k \geq 0} \frac{B_{k,\psi}}{k!}x^k = \frac{x(e^x - e^{2x}-e^{3x}+e^{4x})}{e^{5x}-1} = \frac{xe^x(e^x-1)^2(e^x+1)}{e^{5x}-1} = \frac{2}{5}x^2 - \frac{1}{3}x^4 + \cdots,$$ so looking at the coefficient of $$x^2$$ on both sides tells us $$B_{2,\psi} = 4/5$$.