<s>[INST] Summing various rearrangements of $1-\frac12+\frac13-\frac14+\cdots$ Show that the series $1-\frac12+\frac13-\frac14+\ldots$ converges to $\log2$ but the rearranged series:


*

*$1-\frac12-\frac14-\frac16-\frac18+\frac13-\frac{1}{10}-\frac{1}{12}-\frac{1}{14}-\frac{1}{16}+\frac15-\ldots$ converges to $0$.

*$1+\frac13-\frac12+\frac15+\frac17-\frac14+\ldots$ converges to $\frac32\log2$

*$1+\frac13-\frac12-\frac14+\frac15+\frac17-\frac16+\ldots$ converges to $\log2$

*$1+\frac13+\frac15-\frac12+\frac17+\frac19+\frac{1}{11}-\frac14+\ldots$ converges to $\frac12\log12$
I have absolutely no idea on how to approach these kind of problems. Please help me out from the very start. I consulted books but I didn't understand. I have knowledge on tests for convergence and absolute convergence and the statement of Riemann's theorem. 
 [/INST] Assuming that I have correctly understood the patterns occurring, I get: 
(1)
$$\begin{eqnarray*}\sum_{j=0}^{+\infty}\left(\frac{1}{2j+1}-\frac{1}{8j+2}-\frac{1}{8j+4}-\frac{1}{8j+6}-\frac{1}{8j+8}\right)&=&\sum_{j=0}^{+\infty}\left(-\frac{1}{8j+2}+\frac{3}{8j+4}-\frac{1}{8j+6}-\frac{1}{8j+8}\right)\\&=&\int_{0}^{1}\frac{-x+3x^3-x^5-x^7}{1-x^8}\,dx\\&=&0.\end{eqnarray*}$$
(2)
$$\begin{eqnarray*}\sum_{j=0}^{+\infty}\left(\frac{1}{4j+1}+\frac{1}{4j+3}-\frac{2}{4j+4}\right)&=&\int_{0}^{1}\frac{1+x^2-2x^3}{1-x^4}\,dx\\&=&\frac{3}{2}\log 2.\end{eqnarray*}$$
(3)
$$\begin{eqnarray*}\sum_{j=0}^{+\infty}\left(\frac{1}{4j+1}+\frac{1}{4j+3}-\frac{1}{4j+2}-\frac{1}{4j+4}\right)&=&\int_{0}^{1}\frac{1+x^2-x-x^3}{1-x^4}\,dx\\&=&\log2.\end{eqnarray*}$$
(4)
$$\begin{eqnarray*}\sum_{j=0}^{+\infty}\left(\frac{1}{6j+1}+\frac{1}{6j+3}+\frac{1}{6j+5}-\frac{3}{6j+6}\right)&=&\int_{0}^{1}\frac{1+x^2+x^4-3x^5}{1-x^6}\,dx\\&=&\frac{1}{2}\log 12.\end{eqnarray*}$$
All the integrals can be computed through simple fractions decomposition.
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