Tighter upper bound of $\sum\limits_{k=1}^\infty\frac{\tanh'(\cosh k)}{\cosh(\sin k)}$

Question

How do I find tight upper bounds of $$S=\sum_{k=1}^{\infty}\frac{\tanh'(\cosh k)}{\cosh(\sin k)}$$ ? The derivative of $\tanh$ is $\text{sech}^2$, so using $$S=\sum_{k=1}^\infty\frac{1}{\cosh^2(\cosh k)\cosh(\sin k)}$$we get that the summand is less than $\frac1{k^2}$ so we have $$0<S<\frac{\pi^2}6$$But the upper bound is too big since Desmos states that $S\approx0.122980505644$. How can we make tighter upper bounds? As a note, this sum is already accurate to at least $12$ digits when the upper bound of the sum is $4$.

Edit: I made a tighter bound by comparing this with $\zeta(8)$. So we have $$0<S<\frac{\pi^8}{9450}$$This is still not a tight bound since it is greater than $1$.

Context: I had something similar to this in a dream.

Answer 1

The denominators in $$a_n = \frac{1}{\cosh^2(\cosh k)\cosh(\sin k)}$$ grow extremely fast, so that only few terms are necessary in order to compute $S = \sum_{n=1}^\infty a_n$ to a high precision.

Using $\cosh(\sin k) \ge 1$ and $\cosh k \ge \frac 12 e^k$ we get the estimate $$ \frac{1}{a_k} \ge \left(\frac 12 e^{\frac 12 e^k} \right)^2 = \frac 14 e^{e^k} \, . $$ It follows that $$ a_{n+k} \le 4 e^{-e^{n+k}} = 4 e^{-e^{n}e^{k}} \le 4 \left( e^{-e^n}\right)^k $$ so that the series remainders can be estimated by $$ R_n = \sum_{k=n+1}^\infty a_n = \sum_{k=1}^\infty a_{n+k} \le 4 \frac{e^{-e^n}}{1-e^{-e^n}} = \frac{4}{e^{e^n}-1} \,. $$ So $$ R_3 \le \frac{4}{e^{e^3}-1} \approx 7.549 10^{-9} < 10^{-8} \\ R_4 \le \frac{4}{e^{e^4}-1} \approx 7.769 10^{-24} < 10^{-23} \\ $$ which shows that $a_1+a_2+a_3$ is accurate to at least 8 decimal digits, and $a_1+a_2+a_3+a_4$ is accurate to at least 23 decimal digits. In particular, $$ S \le a_1 + a_2 + a_3 + \frac{4}{e^{e^3}-1} \approx 0.122980513 \, . $$