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.

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    $\begingroup$ Isn't just a rearrangement of $\displaystyle\frac{(-1)^n}{n}$ that converges for Leibniz? $\endgroup$ – edo1998 Feb 23 '18 at 11:09
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    $\begingroup$ Hint: Compute an equivalent of the $k$th block of three consecutive terms. $\endgroup$ – Did Feb 23 '18 at 11:10
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    $\begingroup$ @edo1998 But $\frac{(-1)^n}n$ doesn't converge absolutely, so rearrangements may change the convergence. $\endgroup$ – Arthur Feb 23 '18 at 11:10
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    $\begingroup$ Why is it that 99 out of 100 persons talking about an infinite series wouldn't be able or willing to tell you the definition of convergence, but they will remember (a mutilated version of) the ratio test or the root test? Both tests are (even in their correct form) mostly useless outside the context of power series and their radius of convergence. $\endgroup$ – Professor Vector Feb 23 '18 at 11:11
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    $\begingroup$ @ancientmathematician No, it's a comment, because I'm at work and don't have the time for a real answer. $\endgroup$ – Professor Vector Feb 23 '18 at 12:18

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.

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    $\begingroup$ Also I thought this was a valid reasoning at first, I now came to the conclusion that you are not allowed to group alternating series like that. E.g. by the same reasoning one can argue that $\sum (-1)^n$ converges to zero by grouping $\sum (1+(-1))$. You can group only positive terms, or only negative terms, but not mixed. $\endgroup$ – M. Winter Feb 23 '18 at 11:42
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    $\begingroup$ @M.Winter, if the terms tend to $0$, which they don't for $\sum(-1)^n$, then you are allowed to group a bounded number (in this case three) of terms at a time. $\endgroup$ – Barry Cipra Feb 23 '18 at 11:46
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    $\begingroup$ @BarryCipra I see. Thanks for this enlightenment. I think this is a non-trivial part of the reasoning and should be at least mentioned in the answer. Otherwise one might think this is generally valid (it took me some thinking to get this and I think it needs a proof or a name which can be googled). $\endgroup$ – M. Winter Feb 23 '18 at 11:51
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    $\begingroup$ @M.Winter, see my edited answer. Thanks for the nudge. $\endgroup$ – lhf Feb 23 '18 at 12:10
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    $\begingroup$ @Silent, it's been time since I've looked at how any textbooks present the elementary theory of infinite series, so I can't give any references. Maybe someone else can. As for "bounded number of terms," note that $$1-1+{1\over2}+{1\over2}-{1\over2}-{1\over2}+{1\over3}+{1\over3}+{1\over3}-{1\over3}-{1\over3}-{1\over3}+\cdots$$ fails to converge even though the terms are tending to $0$ and $$(1-1)+({1\over2}+{1\over2}-{1\over2}-{1\over2})+({1\over3}+{1\over3}+{1\over3}-{1\over3}-{1\over3}-{1\over3})+\cdots=0+0+0+\cdots$$ makes it look convergent. $\endgroup$ – Barry Cipra Feb 23 '18 at 13:00

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


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


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.


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 $$ a_n = \begin{cases} \frac{1}{n} & n\ \text{even,}\\ \frac{1}{n+n-1}+ \frac{1}{n+2+n-1} = \frac{1}{2n-1}+\frac{1}{2n+1} & n \ \text{odd,} \end{cases} \qquad a_n \to 0. $$


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

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

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

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

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

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

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.

  • $\begingroup$ I wrote this answer for my memo. Is my answer correct? $\endgroup$ – tchappy ha Jan 28 at 2:16

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