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

Question

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.

Answer 1

My blog post linked in the question gives the key formula which can be used to evaluate the sum in question in closed form.

The relevant formula is $$\vartheta_{2}^{2}(q)=4\sqrt{q}\sum_{n=0}^{\infty}\frac{q^n}{1+q^{2n+1}}\tag{1}$$ It is best to give some details about the key players in the formula above and we start with the basics.

Let $k\in(0,1)$ and $k'=\sqrt{1-k^2}$. The number $k$ is called elliptic modulus and $k'$ is called complementary (to $k$) modulus. Based on these parameters we define complete elliptic integral of first kind $$K(k) =\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}}\tag{2}$$ The numbers $K(k), K(k') $ are usually denoted by $K, K'$ if the corresponding value of $k$ is evident from context.

It is almost magical that given the values of $K, K'$ the value of $k$ can be obtained via functions of a variable $q=e^{-\pi K'/K} $ (called nome corresponding to $k$). Let's then define Jacobi's theta functions \begin{align} \vartheta_{2}(q)&=\sum_{n\in\mathbb{Z}}q^{(n+(1/2))^2}\tag{3a}\\ \vartheta_{3}(q)&=\sum_{n\in\mathbb{Z}}q^{n^2}\tag{3b} \end{align} and then we have the identities $$\vartheta_{3}^{2}(q)=\frac{2K}{\pi}, \, k=\frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)}\tag{4}$$ The formula $(1)$ and the above identity $(4)$ can be proved using definitions $(2),(3a),(3b)$ but the proofs are non-trivial and non-obvious.

---

Coming back to the sum in question, let's write $q=e^{-\pi} $ and the sum in question is equal to $$\sum_{n=0}^{\infty}\frac{2}{q^{2n+1}+q^{-(2n+1)}}=2q\sum_{n=0}^{\infty}\frac{q^{2n}}{1+q^{4n+2}}=\frac{\vartheta_{2}^{2}(q^2)}{2}\tag{5}$$ From $(4)$ we have $$\vartheta_{2}^{2}(q)=\frac{2kK}{\pi}$$ Next we have Landen transformation which allows us to replace $q$ by $q^2$ and accordingly replace $k$ by $(1-k')/(1+k')$ and $K$ by $(1+k')K/2$. And therefore we have $$\vartheta_{2}^{2}(q^2)=\frac{(1-k')K}{\pi}\tag{6}$$ Now using $(5),(6)$ we get our sum as $(1-k')K/(2\pi)$. For $q=e^{-\pi} $ we have $K=K'$ so that $k=k'=1/\sqrt{2}$ and $K=\dfrac{\Gamma^2(1/4)}{4\sqrt{\pi}}$ and we get the desired closed form evaluation of the sum in question.

Answer 2

$$ \eqalign{\text{sech}((2n-1)\pi) &= \frac{2}{\exp((2n-1)\pi) + \exp(-(2n-1)\pi)}\cr &= \frac{2 \exp(-(2n-1)\pi)}{1 + \exp(-2(2n-1)\pi)}\cr &= 2 \sum_{j=0}^\infty (-1)^{j} \exp(-(2j+1)(2n-1)\pi)}$$ 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 $n \equiv 3 \mod 4$.

Answer 3

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.

Answer 4

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.