# Probably this :$\sum_{n=2}^{\infty}(-1)^n\frac{\arctan{(1-2^n)}\log n \tan{(1-2^{-n})}}{n^3\sqrt{n}\log \log n }$ is Euler constant

I'm always interesting to find some approach in the form of series or integral to get any known constant , In this once i have accrossed in my mind to use some trigonometrics functions in the form of Altern series as shown below , My computation in wolfram alpha for some large $n$ showed to me that is close or coincide with Euler constant with approximation of $10^{-3}$, The formula $A$ is complicated for evaluation using frobinus integral which seems suitable for that but i think no luck with it , Now my qustion here how i can evaluate that sum $A$

$$\sum_{n=2}^{\infty}(-1)^n\frac{\arctan{(1-2^n)}\log n \tan{(1-2^{-n})}}{n^3\sqrt{n}\log \log n }\tag{A}$$ ?

Note I meant Euler-Mascheroni constant.

• O_o ... nani??? Commented Aug 25, 2018 at 21:00

Instead it converges to some number starting with the digits 0.5722... To see this, notice that $$\frac{\arctan{(1-2^n)}\log n \tan{(1-2^{-n})}}{n^3\sqrt{n}\log \log n }$$ is decreasing when $n>3$. This means that the alternating sum wil lie in between two consecutive partial sums. The 30th partial sum is 0.57225... and the 31th is 0.57229...