# Evaluate infinite hyperbolic series $\sum_{n=1}^{\infty} \mathrm{sech}((2n-1)\pi)$

Could anyone show me a solution how to approach this series? According to this blog I doubt that we could be using $$\theta_{2}^{2}(q^2)=4\sum_{n=1}^{\infty} \frac{q^{(2n+1)}}{1+q^{(4n+2)}}$$ and let $$\frac{K’}{K}=1$$ So $$k$$ and $$k’ =\frac{1}{\sqrt{2}}$$.

But I don’t know about Theta function much ,what’s related to $$\theta_{2}^{2}(q^2)~,$$ and I’m quite new to this function.Any help would be very appreciated.

• Numerics suggests that it is approximately equal to $\frac{2e^{-\pi}}{1+e^{-2\pi}}.$ – Dr Zafar Ahmed DSc Aug 14 at 4:24
• @DrZafarAhmedDSc that is simply because your value equals $\operatorname{sech}(\pi)$, the first term in this series, and $\operatorname{sech}(x)$ rapidly decays so the first term is dominant. – Brevan Ellefsen Aug 14 at 4:47

## 3 Answers

I'll sketch why there is a closed-form for those kind of special values of modular forms.

• From that $$sech(\pi t)$$ is its own Fourier transform, with the Poisson summation formula show that $$f(z) = \sum_n sech(2i \pi z)$$ is a modular form of weight $$1$$ for $$\Gamma_0(4)$$.

• We want to evaluate $$f(i/2)$$, equivalently $$f(8i)$$ using the modularity. Let $$g(z) = \frac{f(8z)^{12}}{\Delta( z)}$$, it is weight $$0$$ modular of level $$4.8$$ and it is in the function field of $$X_0(32)$$, with no poles on the upper-half plane and it has integer coefficients. Thus $$P(z,T) = \prod_{d| 32, b \bmod d} (T-g(\frac{\frac{32}d z+b}{d})) = p(j(z),T)$$ with $$p \in \Bbb{Q}[X,Y]$$. The coefficients of that polynomial are found from the first few coefficients of $$g(\frac{\frac{32}d z+b}{d})$$ at $$i\infty$$.

• Together with $$j(i) = 1728$$ we find that $$g(i)$$ is a root of $$p(1728,T)\in \Bbb{Q}[T]$$. We can factorize this polynomial to find the minimal polynomial of $$g(i)$$.

• Finally from the isomorphism given by Weierstrass functions from complex tori to elliptic curves we find $$\Delta(i)$$ is given in term of the integral $$\int_C \frac{dx}{y}=\int_\infty^0 \frac{1}{\sqrt{x^3+x}}dx+\int_0^\infty \frac{1}{-\sqrt{x^3+x}}dx$$ on the elliptic curve $$y^2= x^3+x$$, and the latter integrals reduce to the $$\beta(.,.)$$ function and to $$\Gamma(1/4)$$, obtaining that $$\Delta(i ) = (2\pi)^{-6} \Gamma (1/4)^{24}$$.

Whence we have found the minimal polynomial of $$f(8i)^{12} (2\pi)^6 \Gamma(1/4)^{-24}$$ and a radical expression for it, ie. a closed-form, because that polynomial is radical.

• From there you can try to implement the needed algorithm to find the polynomial and factorize it, but you'll probably have some troubles because it is large, thus you'll need some tables of modular forms and their particular values (in order to split the computation into several simpler steps with smaller polynomials). – reuns Aug 14 at 16:41

\eqalign{\text{sech}((2n-1)\pi) &= \frac{2}{\exp((2n-1)\pi) + \exp(-(2n-1)\pi)}\cr &= \frac{2 \exp(-(2n-1)\pi)}{1 + \exp(-2(2n-1)\pi)}\cr &= 2 \sum_{j=0}^\infty (-1)^{j} \exp(-(2j+1)(2n-1)\pi)} so that \eqalign{\sum_{n=1}^\infty \text{sech}((2n-1)\pi) &= 2 \sum_{n=1}^\infty \sum_{j=0}^\infty (-1)^j \exp(-(2j+1)(2n-1)\pi)\cr &= 2 \sum_{m=0}^\infty f(2m+1) \exp(-(2m+1)\pi)} where $$f(n)$$ is the number of divisors of $$n$$ that are congruent to $$1$$ mod $$4$$ minus the number of divisors congruent to $$3$$ mod $$4$$. Note that $$f(n)=0$$ if $$n \equiv 3 \mod 4$$.

According to a CAS $$\sum_{n=1}^{\infty} \text{sech}((2n-1)\pi)=-\frac{\pi+i \left(\psi _{e^{2 \pi }}\left(\frac{2+i}{4}\right)- \psi _{e^{2 \pi }}\left(\frac{2-i}{4}\right)\right) }{2 \pi }$$ where appears the q-digamma function.