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

2. $1+\frac13-\frac12+\frac15+\frac17-\frac14+\ldots$ converges to $\frac32\log2$

3. $1+\frac13-\frac12-\frac14+\frac15+\frac17-\frac16+\ldots$ converges to $\log2$

4. $1+\frac13+\frac15-\frac12+\frac17+\frac19+\frac{1}{11}-\frac14+\ldots$ 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.

• It would probably be helpful if instead of truncating the series' by hand you wrote them out in a more standardized notation so we can easily see the pattern being displayed. – Ethan Aug 27 '14 at 19:25
• @Diya The wikipedia article on the Riemann series theorem shows how to find rearrangements of the series for $\ln(2)$ & the constants they approach. en.wikipedia.org/wiki/… – Jam Aug 27 '14 at 19:34
• I guess the first one is repeatedly +1 odd, -4 evens, ... The second is +2 odds, -1 even, ... The third +2 odds, -2 evens, ... The fourth +3 odds, -1 even, ... – punctured dusk Aug 27 '14 at 19:36
• @Ethan that's another problem, I can't really understand the pattern here. – Diya Aug 27 '14 at 19:46

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.

• Maybe you should explain to the OP how you switch from sums to integrals? – Ali Caglayan Aug 27 '14 at 20:35
• $$\int_{0}^{1}x^m\,dx = \frac{1}{m+1}.$$ – Jack D'Aurizio Aug 27 '14 at 20:36
• If we have a rearrangement of the series which diverges, will the integral you get diverge also in that case? – user84413 Aug 27 '14 at 22:57

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.