Verified answer

Others have given simple reasons why it converges. I find the sum is equal to

∞

∑(-1)^(n+1)*(1/(2n+1))*ζ(2n+1) , where ζ is the Riemann Zeta function given by ζ(2n+1) =

n=1

∞

∑1/k^(2n+1).

k=1

Since, so far as I know, explicit formulas are not known for ζ(odd integer), this makes me pessimistic that a closed form value can be given for this sum. The approximate numerical value (using maple) is:

0.27557534443399966

Converges by integral test.

I haven't proved the sum yet.

The sum is 0.275 ( I use a computer software )

It converges since lim{n->inf} [1/n - arctan(1/n)]/(1/n^2)] = 0

sum = ∫[1,0] (x - arctan(x)) dx