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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.

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    $\begingroup$ O_o ... nani??? $\endgroup$ Commented Aug 25, 2018 at 21:00

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This sum does not converge to the Euler-Mascheroni constant.

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...

This means that the actual value of the infinite sum wil be 0.5722... where the next digit is between 5 and 9. The Euler-Mascheroni constant instead starts with 0.5772...

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