The harmonic alternative

Split the series into two parts: the even denominators and the odd denominators.

The even denominators is just the regular alternating harmonic series divided by 2, so this is just \frac{ln2}{2}.

The odd denominators are a little trickier. Start with the series \frac{1}{1+x^2} = 1-x^{2}+x^{4}-x^{6}...

Then take the integral of both sides. Tan inverse of (x) = x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}....

Plug in x=1 to get: Tan inverse of (1) = 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}....

\frac{\pi}{4} = the series of odd denominators.

So the answer is \frac{\pi}{4} + \frac{ln2}{2}, which is about \boxed{1.132}

Let the given infinite summation be $S$ . Now consider the Maclaurin series of $\ln (1-i)$ , where $i=\sqrt{-1}$ is the imaginary unit .

$\begin{aligned} \ln (1-i) & = - \frac i1 + \frac 12 + \frac i3 - \frac 14 - \frac i5 + \frac 16 + \frac i7 - \frac 18 + \cdots & \small \color{#3D99F6} \text{Multiply both sides by }i \\ i \ln (1-i) & = \frac 11 + \frac i2 - \frac 13 - \frac i4 + \frac 15 + \frac i6 - \frac 17 - \frac i8 + \cdots \end{aligned}$

$\implies S = \Re (i \ln (1-i)) + \Im (i \ln (1-i))$ , where $\Re(z)$ and $\Im(z)$ are real part and imaginary part of complex number $z$ respectively.

$\begin{aligned} i\ln (1- i) & = i \ln \left(\sqrt 2 \color{#3D99F6} \left(\frac 1{\sqrt 2} - \frac i{\sqrt 2} \right) \right) & \small \color{#3D99F6} \text{By Euler's formula: }e^{\theta i} = \cos \theta + i\sin \theta \\ & = i \ln \left(\sqrt 2 \color{#3D99F6} e^{-\frac \pi 4 i} \right) \\ & = i \left(\frac {\ln 2}2 - \frac \pi 4 i \right) \\ & = \frac \pi 4 + \frac {\ln 2}2 i \end{aligned}$

Therefore, $S = \Re (i \ln (1-i)) + \Im (i \ln (1-i)) = \dfrac \pi 4 + \dfrac {\ln 2}2 \approx \boxed{1.132}$ .