Finding out $S:=1+\frac12-\frac13-\frac14+\frac15+\frac16-\frac17-\frac18+\cdots$

Solution 1:

Note that $$S=A+B$$ where $\displaystyle A=\sum_{k\ge 1}\frac{(-1)^{k-1}}{2k-1},\ B=\sum_{k\ge 1}\frac{(-1)^{k-1}}{2k}$. Clearly, $B=\displaystyle \frac{\ln 2}{2}$. and, $A=\tan^{-1}1=\pi/4$ which you can find out using your technique as below $$A=\sum_{k\ge 1}(-1)^{k-1}\int_0^1 x^{2k-2}dx\\=\int_{0}^1\sum_{k\ge 1}(-1)^{k-1}x^{2k-2}dx\quad(\mbox{Use Fubini to justify the change of order})\\=\int_{0}^1 \frac{dx}{1+x^2}=\tan^{-1}1=\pi/4\\ \mbox{similarly, }\ B=\int_{0}^1 \sum_{k\ge 1}(-1)^{k-1}x^{2k-1}dx\\=\int_0^1 \frac{x}{1+x^2}dx\\=\frac{\ln 2}{2}$$

Solution 2:

Couldn't you just write this as

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} + \frac12 \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k+1} = \frac{\pi}{4} + \frac12 \log{2}$$