I was investigating the asymptotic properties of the $\cosh$ functions and how they all strongly relate to $e^x$
In my studying, I found out that $\cosh x\sim \frac{e^x}{2}$
By that definition, that must also mean that $\cosh^{-1} x\sim \log 2x$
From that finding, I wanted to a bit deeper into the approximation, but without the use of Big Oh's.
That led me to the sum of my question:
What is the closed form to $$I=\sum_{n=1}^\infty \left(1- \frac{\cosh^{-1} n}{\log 2n}\right)$$
W|A could not evaluate the sum, but did come up with an approximation to it. That approximation is $I=1.09258$