MHB Summation Challenge #1: Evaluate $\sum$

[Saitama]

Evaluate the following:
$$\Large \sum_{k=1}^{\infty} (-1)^{\left\lfloor \frac{k+3}{2} \right\rfloor} \frac{1}{k}$$

 

[Opalg]

[sp]$$\begin{aligned}\sum_{k=1}^{\infty} (-1)^{\left\lfloor \frac{k+3}{2} \right\rfloor} \frac{1}{k} &= 1 + \frac12 - \frac13 - \frac14 + \frac15 + \frac16 - \frac17 -\frac18 + \ldots \\ &= \Bigl(1 - \frac13 + \frac15 - \frac17 + \ldots\Bigr) + \frac12\Bigl(1 - \frac12 + \frac13 - \frac14 + \ldots\Bigr) \\ &= \frac\pi4 + \frac12\ln2\end{aligned}$$[/sp]

 

[Saitama]

[sp]$$\begin{aligned}\sum_{k=1}^{\infty} (-1)^{\left\lfloor \frac{k+3}{2} \right\rfloor} \frac{1}{k} &= 1 + \frac12 - \frac13 - \frac14 + \frac15 + \frac16 - \frac17 -\frac18 + \ldots \\ &= \Bigl(1 - \frac13 + \frac15 - \frac17 + \ldots\Bigr) + \frac12\Bigl(1 - \frac12 + \frac13 - \frac14 + \ldots\Bigr) \\ &= \frac\pi4 + \frac12\ln2\end{aligned}$$[/sp]

Great as always! :)