Summing various rearrangements of $1-\frac12+\frac13-\frac14+\cdots$

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.

This is more sketch than proof, but I hope it gets the point across: Once you've got the first result (the alternating sum converges to $\ln2$), the others follow by examining partial sums.

For 1., notice that the partial sums, truncated every fifth term (i.e., just before the next odd reciprocal is added), can be written as

$$S=\left(1+{1\over3}+\cdots+{1\over2n-1} \right)-\left({1\over2}+{1\over4}+\cdots+{1\over8n} \right)$$

Let's write the familiar sum $\ln2=1-{1\over2}+{1\over3}-{1\over4}+\cdots$ as

$$\ln2\approx\left(1+{1\over3}+\cdots+{1\over2n-1} \right)+\left({1\over2n+1}+\cdots+{1\over8n-1} \right)-\left({1\over2}+{1\over4}+\cdots+{1\over8n} \right)$$

Thus

$$S\approx\ln2-\left({1\over2n+1}+\cdots+{1\over8n-1} \right)$$

But

$${1\over2n+1}+\cdots+{1\over8n-1}\approx\int_n^{4n}{1\over2x+1}dx={1\over2}\ln\left({8n+1\over2n+1}\right)\approx\ln2$$

Thus

$$S\approx\ln2-\ln2=0$$

The "$\approx$" signs become "$=$" in the limit as $n\to\infty$. The other cases, I believe, can be handled similarly.