Convergence of $1+\frac13-\frac12+\frac15+\frac17-\frac14+\frac19+\frac1{11}-\frac16+\ldots$

Question

I was reading Rudin PMA, example 3.53 on P76. There he points out that rearrangement may not give same limit of a series. Then he says that it is left as exercise to show that above mentioned series converges. I thought clubbing three terms, but did not seem 'legal'.

How to show that series $1+\dfrac13-\dfrac12+\dfrac15+\dfrac17-\dfrac14+\dfrac19+\dfrac1{11}-\dfrac16+\ldots$ converges?

I tried root test: Let this series be $\sum_{n=1}^{\infty}a_n$, then $\limsup\limits_{n\to\infty}\sqrt[n]{|a_n|}=1$, since basically, this series is rearrangement of $\sum \frac {(-1)^n}n$. So root test is inconclusive.

Answer 1

In blocks of $3$ terms, the series is $$ \sum_{n=0}^{\infty} \left(\dfrac{1}{4n+1}+\dfrac{1}{4n+3}-\dfrac{1}{2n+2}\right) = \sum_{n=0}^{\infty} \dfrac{8 n + 5}{32 n^3 + 64 n^2 + 38 n + 6} \le \dfrac56+ \sum_{n=1}^{\infty} \dfrac{1}{n^2} < \infty $$

This proves that the partial sums $S_{3n+2}$ of the original series converge. The other partial sums, $S_{3n+1}$ and $S_{3n}$, differ from $S_{3n+2}$ by one or two terms that converge to zero and so they converge as well, to the same limit. Thus, the partial sums converge, that is, the series converges.

Answer 2

It seems like the series is alternatingly composed of positive terms (and it is okay to group them) of the form

$$a_{2n}=\frac1{4n+1}+\frac1{4n+3}=\frac{8n+4}{(4n+1)(4n+3)}=\frac{8n+4}{16n^2+16n+3}=\frac{n+1/2}{2n^2+2n+3/8}.$$

for $n=0,1,2,...$, and negative terms of the form

$$a_{2n+1}=-\frac{1}{2n+2}.$$

also for $n=0,1,2,...$. Note that for sufficiently large $n$ we have

$$ \underbrace{\frac1{2n}}_{|a_{2n-1}|} \ge \underbrace{\frac{n+1/2}{2n^2+2n+3/8}}_{|a_{2n}|} \ge \underbrace{\frac 1{2n+2}}_{|a_{2n+1}|}. $$

So we have an alternating series with absolutely decreasing terms with $|a_n|\to 0$. We can apply the alternating series test to reason that the sum converges.

Answer 3

As many people already pointed out, the series converges thanks to Leibniz criterion! Indeed the series is $$ \sum_{n=1}^\infty (-1)^{n+1} a_n, $$ where $a_n$ are $$ a_n = \begin{cases} \frac{1}{n} & n\ \text{even,}\\ \frac{1}{n+n-1}+ \frac{1}{n+2+n-1} = \frac{1}{2n-1}+\frac{1}{2n+1} & n \ \text{odd,} \end{cases} \qquad a_n \to 0. $$

Answer 4

$c_1 := 1+\frac{1}{3}$.
$c_2 := -\frac{1}{2}$.
$c_3 := \frac{1}{5} + \frac{1}{7}$.
$c_4 := -\frac{1}{4}$.
$c_5 := \frac{1}{9} + \frac{1}{11}$.
$c_6 := -\frac{1}{6}$.
$\cdots$
$c_{2 k - 1} := \frac{1}{4 k - 3} + \frac{1}{4 k - 1}$ for $k \in \{1, 2, 3, \cdots\}$.
$c_{2 k} := -\frac{1}{2 k}$ for $k \in \{1, 2, 3, \cdots\}$.

Then,
(a)
$\frac{1}{4 k - 3} + \frac{1}{4 k - 1} > \frac{1}{4 k} + \frac{1}{4 k} > \frac{1}{4 k + 1} + \frac{1}{4 k + 3}$.
$\therefore |c_{2 k - 1}| > |c_{2 k}| > |c_{2 k + 1}|$ for $k \in \{1, 2, 3, \cdots\}$.

(b)
$c_{2 k - 1} \geq 0$ and $c_{2 k} \leq 0$.

(c)
$\lim_{k\to\infty} c_{2 k - 1} = \lim_{k\to\infty} c_{2 k} = 0$.
$\therefore \lim_{n\to\infty} c_n = 0$.

So, $\sum c_n$ is an alternating series and converges by Theorem 3.43 on p.71.

$c_1 = s'_2$.
$c_1 + c_2 = s'_3$.
$c_1 + c_2 + c_3 = s'_5$.
$c_1 + c_2 + c_3 + c_4 = s'_6$.
$c_1 + c_2 + c_3 + c_4 + c_5 = s'_8$.
$c_1 + c_2 + c_3 + c_4 + c_5 + c_6= s'_9$.
$\cdots$

So, $s'_2, s'_3, s'_5, s'_6, s'_8, s'_9, \cdots$ is a convergent subsequence of $\{s'_n\}$.
So, $s'_2, s'_5, s'_8, s'_{11}, \cdots$ is a convergent subsequence of $\{s'_n\}$.
$s'_{3 k - 2} = s'_{3 k - 1} - \frac{1}{4 k - 1}$.
So, $s'_1, s'_4, s'_7, s'_{10}, \cdots$ is a convergent subsequence of $\{s'_n\}$.
$\{s'_1, s'_4, s'_7, s'_{10}, \cdots\} \cup \{s'_2, s'_3, s'_5, s'_6, s'_8, s'_9, \cdots \} = \{s'_1, s'_2, s'_3, s'_4, s'_5, s'_6, s'_7, s'_8, s'_9, s'_{10}, \cdots\}$.
So, $\{s'_n\}$ converges.