I ran across this series involving arctan and I am very curious to know if it has a closed form.
$$ \sum_{k=1}^{\infty} \frac{1}{k}\arctan{\frac{1}{k}} $$
It looks too devilishly simple to not have a wonderful closed form. For a little context a series like this will arise from integrating the relationship: $$ \sum_{k=1}^{\infty}\frac{1}{k^2+x^2}=\frac{\pi\coth\pi x}{2x}-\frac{1}{2x^2} $$ Thanks for any help!
$$S=\sum_{k=1}^{\infty} \frac{1}{k}\tan ^{-1}\left(\frac{1}{k}\right)$$
Using $$\frac{1}{k}\tan ^{-1}\left(\frac{1}{k}\right)=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2 n-1}\frac 1 {k^{2n}}$$ Changing the order of summation $$S=\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\zeta (2 n)}{2 n-1}=1.405869298287780\cdots$$ This number is not recognized by inverse symbolic calculators.