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

Question

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.

Answer 1

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.

Answer 2

Here is an elementary proof of the general result.

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Suppose the alternating harmonic series is rearranged into blocks of $a$ positive (odd-denominator) terms and $b$ negative (even-denominator) terms. For example, series 1 in the OP has $a=1$, $b=4$; the usual alternating harmonic series has $a=b$, as does series 3 in the OP.

Claim: The rearrangement converges to $\log 2+\frac12\log(\frac ab)$.

Proof: The sum of the first $n$ blocks is $$ \underbrace{\left(\frac11+\frac13+\cdots+\frac1{2an-1}\right)}_{\text{$an$ odd denominators}} -\underbrace{\left(\frac12+\frac14+\cdots+\frac1{2bn}\right)}_{\text{$bn$ even denominators}}.\tag1 $$ We recognize the right member of (1) as $\frac12 H_{bn}$, where $H_k$ denotes the partial harmonic sum $\frac11+\frac12+\cdots+\frac1k$, while the left member equals $H_{2an}-\frac12H_{an}$. So (1) can be written $$ H_{2an}-\frac12H_{an}-\frac12H_{bn}= (H_{2an}-H_{an}) +\frac12(H_{an}-H_{bn}).\tag2 $$ Let $n\to\infty$. Apply this result to see that $H_{2an}-H_{an}\to\log 2$ and $H_{an}-H_{bn}\to\log(a/b)$.

This shows that the blocked partial sums of the rearrangement converge to the claimed limit. The unblocked series converges to the same limit, since its terms tend to zero and the blocks have the same size.

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The same reasoning proves the more general case where the blocks have variable size, such that the first $n$ blocks consume $a_n\ge n$ positive terms and $b_n\ge n$ negative terms of the AHS. If the blocks remain bounded in size and $a_n/b_n\to c>0$, then the rearranged series converges to $\log 2 +\frac12\log c$. If $a_n/b_n\to \infty$, the rearranged series diverges to $\infty$.

Answer 3

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.