I was trying to find closed form generalizations of the following well known hyperbolic secant sum $$ \sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n}=\frac{\left\{\Gamma\left(\frac{1}{4}\right)\right\}^2}{2\pi^{3/2}},\tag{1} $$ as $$ S(a)=\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+a}. $$ In particular I find by numerical experimentation $$ \displaystyle \frac{\displaystyle\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}}{\displaystyle\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n}}\overset{?}=-\frac{1}{2}\left(1+\sqrt{2}\right)+\sqrt{2+\sqrt{2}}\tag{2} $$ (Mathematica wasn't able to find a closed form directly, but then I decided to switch to calculation of ratios of the sums, calculated ratios numerically and then was able to recognize this particular ratio as a root approximant. This was subsequently verified to 1000 decimal places).

I simplified this expression from the previous edition of the question.

Unfortunately for other values of $a$ I couldn't find a closed form. Of course $(2)$ together with $(1)$ would imply a closed form for the sum $S(1/\sqrt{2})$

How one can prove $(2)$?

  • 6
    $\begingroup$ You'd have to give us more than "hey, this expression is close" to give us a reason to even start believing the conjecture. How about tell us the exact value you got from numerically evaluation? $\endgroup$ – Thomas Andrews Jan 16 '16 at 14:37
  • 6
    $\begingroup$ I've a conjectured value for $S(1/2)$. $$\sum_{n=-\infty}^\infty\frac1{\cosh\pi n + 1/2}\stackrel{?}{=} \sqrt{\frac23\sqrt{72+42\sqrt3}-2\sqrt3} \cdot \frac{\Gamma^2\left(\frac14\right)}{6\pi^{3/2}}$$ $\endgroup$ – user153012 Jan 18 '16 at 0:02
  • 2
    $\begingroup$ Check out Andreas Dieckmann's table, under "Series of Hyperbolic Functions". $\endgroup$ – nospoon Jan 19 '16 at 15:33
  • $\begingroup$ Using properties of symmetry of the $e^{πn}$ function and the definition of the hyperbolic cosine, this sum can be rewritten as: $4\sum_{n=0}^\infty\frac{e^{πn}}{e^{2πn}+1+\sqrt2 e^{πn}}$, Which passes the comparison test and does converge. WolframAlpha's approximation yields 0.335494, and that website provides a partial fraction formula $\endgroup$ – KR136 Jan 24 '16 at 22:21
  • $\begingroup$ @KaR1367 thanks for your analysis. Can you show that the complicated expression you obtained equals $-\frac{1}{2}\left(1+\sqrt{2}\right)+\sqrt{2+\sqrt{2}}$ when $m\to \infty$? $\endgroup$ – user294724 Jan 26 '16 at 15:38

Let $$ S_1(\alpha)=\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi \alpha n+\frac{1}{\sqrt{2}}}, $$ $$ S_2(\alpha)=\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi \alpha n-\frac{1}{\sqrt{2}}}, $$ then due to $2\cosh^2x-1=\cosh 2x$ one obtains $$ S_2(\alpha)-S_1(\alpha)=2\sqrt{2}\sum_{n=-\infty}^\infty\frac{1}{\cosh 2\pi \alpha n}, $$ $$ S_2(\alpha)+S_1(\alpha)=4\sum_{n=-\infty}^\infty\frac{\cosh\pi\alpha n}{\cosh 2\pi \alpha n}. $$ Now if one defines elliptic integrals of the first kind $K$ and $\Lambda$ according to equations $\frac{K'}{K}=\frac{K(k')}{K(k)}=\alpha$, $\frac{\Lambda'}{\Lambda}=\frac{K(k_1')}{K(k_1)}=2\alpha$, where $k'=\sqrt{1-k^2},~k_1'=\sqrt{1-k_1^2}$, then the well known formulas from the theory of elliptic functions (see Whittaker and Watson, A Course of Modern Analysis) state that $$ \sum_{n=-\infty}^\infty\frac{1}{\cosh \pi \alpha n}=\frac{2K}{\pi},~\sum_{n=-\infty}^\infty\frac{1}{\cosh 2\pi \alpha n}=\frac{2\Lambda}{\pi},~\sum_{n=-\infty}^\infty\frac{\cosh\pi\alpha n}{\cosh 2\pi \alpha n}=\frac{2\Lambda}{\pi}~\text{dn}(i\Lambda'/2,k_1), $$ $$ k_1=\frac{1-k'}{1+k'},\quad \Lambda=\frac{1}{2}(1+k')K,\quad \text{dn}(i\Lambda'/2,k_1)=\sqrt{1+k_1}. $$ From this by trivial algebra one can deduce that

$$ S_1(K'/K)=\frac{K\sqrt{2}}{\pi}(1+k')\left(\frac{2}{\sqrt{1+k'}}-1\right). $$

Now for $k=1/2$ one has $k'=1/2$, $K=K'=K_0$, therefore $$ \frac{\displaystyle\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}}{\displaystyle\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n}}=\frac{S_1(1)}{2K_0/\pi}=\frac{(1+k')}{\sqrt{2}}\left(\frac{2}{\sqrt{1+k'}}-1\right)=\sqrt{2+\sqrt{2}}-\frac{1+\sqrt{2}}{2}. $$


Although I will not give you any solution, nevertheless I will try write an attempt.

I shall start with observation - the value $1/\sqrt{2}$ is rather peculiar for $\arccos$ since $\arccos\frac{1}{\sqrt{2}}=\frac{\pi}{4}$ ...

Define a meromorphic function $$f(z):=\frac{\cot\pi z}{\cosh \pi z + a}$$

Integrating this function along rectangular contour one simply gets the relation :

$$\sum_{n=-\infty}^{\infty}\frac{1}{\cosh\pi n+a}=-2\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{\cosh\pi x-a}-\frac{2}{\sqrt{1-a^2}}\sum_{n=-\infty}^{\infty}\frac{\sinh(2\arccos(-a))}{\cosh 4\pi n-\cosh(2\arccos(-a))}$$

For $a=1/\sqrt{2}$ the second sum is according to Mathematica equal :

$$\frac{Q+2 \pi \coth \left(\frac{3 \pi }{4}\right)-3\pi}{2 \pi\sinh \left(\frac{3 \pi }{2}\right) }$$

where $Q$ is a constant expressible in terms of Q-Gamma function :

$$Q=\psi _{e^{4 \pi }}\left(\frac{3}{8}\right)-\psi _{e^{4 \pi }}\left(-\frac{3}{8}\right)$$

The integral was calculated already by the user @sirfoga, in our case for the first term : $$-2\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{\cosh\pi x-a}=-\frac{4\arctan{\left(\frac{1+a}{\sqrt{1-a^2}}\right)}}{\pi\sqrt{1-a^2}}$$

For the special case of $a=1/\sqrt 2$ is this equal to $-3/\sqrt 2$ so for the overall sum


Connection to the OP's conjecture is at this stage barelly visible :/

Addendum I have run the numbers on Mathematica and it seems the conjecture holds at least for 100 000 digits

  • 1
    $\begingroup$ The main problem with this answer lies in the fact that it rewrites a mysterious series of a certain form in terms of another series of the same form. But if one is aware of a mechanism to evaluate the latter, then that mechanism could have been used to express the former directly. $\endgroup$ – Lucian Aug 23 '16 at 15:29

Disclaimer: I verified my self that $\frac{\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}}{\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n}} \approx 0.640652 \approx \sqrt{\sqrt{2}+2}-\frac{1}{2} \left(\sqrt{2}+1\right)$ so mine is only an approximation but as far as I could do, no closed form exists to evaluate precisely your sum: besides, if your aim is to do computational evaluations, I suggest to simplify the sum, evaluate the integral I'm providing or evaluate the sum from $n = -10^3$ to $n = 10^3$: I've seen no massive difference with $n = -10^6$ to $n = 10^6$ or more.

\begin{align} \sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}} &\approx \int_{-\infty}^\infty\frac{1}{\cosh\pi x+\frac{1}{\sqrt{2}}} dx\\ &= -\frac{2 \sqrt{2} \arctan\left(\frac{(\sqrt{2}-2) \tanh \left(\frac{\pi n}{2}\right)}{\sqrt{2}}\right)}{\pi } \big|_{-\infty}^{+\infty}\\ &= \sqrt{2} \end{align} and more in general \begin{align} \sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+a} &\approx \int_{-\infty}^\infty\frac{1}{\cosh\pi x+a} dx\\ &= -\frac{2 \arctan\left(\frac{(a-1) \tanh \left(\frac{\pi n}{2}\right)}{\sqrt{1-a^2}}\right)}{\pi \sqrt{1-a^2}} \big|_{-\infty}^{+\infty}\\ &= \left( -\frac{2 \arctan\left(\frac{a-1}{\sqrt{1-a^2}}\right)}{\pi \sqrt{1-a^2}} \right) - \left(\frac{2 \arctan\left(\frac{a-1}{\sqrt{1-a^2}}\right)}{\pi \sqrt{1-a^2}} \right)\\ &= -2 \left(\frac{2 \arctan\left(\frac{a-1}{\sqrt{1-a^2}}\right)}{\pi \sqrt{1-a^2}} \right) \end{align} so in your specific case \begin{align} \frac{\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}}{\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n}} &\approx \frac{\int_{-\infty}^\infty\frac{1}{\cosh\pi x+\frac{1}{\sqrt{2}}} dx}{\int_{-\infty}^\infty\frac{1}{\cosh\pi x} dx}\\ &= \frac{-2 \left(\frac{2 \arctan\left(\frac{\frac{1}{\sqrt{2}}-1}{\sqrt{1-\frac{1}{\sqrt{2}^2}}}\right)}{\pi \sqrt{1-\frac{1}{\sqrt{2}^2}}} \right)}{-2 \left(\frac{2 \arctan\left(\frac{0-1}{\sqrt{1-0^2}}\right)}{\pi \sqrt{1-0^2}} \right)}\\ &= \frac{1}{\sqrt{2}}\\ &\approx 0.707107 \end{align}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy