I am trying to find a way to prove that $$\dfrac 11 + \dfrac 12 + \dfrac 13 + \dfrac 14 + \cdots \color{red}{-} \dfrac 18 + \cdots$$

where the pattern repeats every $8$ terms. Knowing about the Riemann Series Theorem, I am a little hesitant about manipulating conditionally convergent series at all. Granted that the harmonic series diverges, is the following a valid way to prove my series diverges?

$$\dfrac 11 + \dfrac 12 + \dfrac 13 + \cdots + \dfrac 17 - \dfrac 18 + \cdots = \sum_{n=1}^{\infty} \dfrac 1n - 2 \cdot \dfrac 18\sum_{n=1}^{\infty} \dfrac 1n $$

$$=\dfrac 34 \sum_{n=1}^{\infty} \dfrac 1n$$

Since the harmonic series diverges, so does $\dfrac 34$ times it.

  • 4
    $\begingroup$ I stared at that line many, many times before I finally saw where the $+$ turned into a $-$. $\endgroup$ – Yakk Oct 21 '16 at 10:55
  • $\begingroup$ @Yakk I suggested an edit to make it more obvious $\endgroup$ – null Oct 21 '16 at 16:46
  • $\begingroup$ I've made it red. Don't know if there's a better way. $\endgroup$ – asmeurer Oct 21 '16 at 20:11
  • $\begingroup$ @asmeurer Is it better now? $\endgroup$ – Ovi Oct 22 '16 at 1:57
  • $\begingroup$ That helps. At first I thought you meant repeat the terms shown, which obviously diverges. $\endgroup$ – asmeurer Oct 22 '16 at 1:59

Your idea is a good one, but, as you suspected, you need to be more careful about this sort of manipulation of conditionally convergent series.

One way to carry out your argument correctly, but with only minor changes, is by looking at partial sums:

Let's write $$a_n=\begin{cases}1,&\text{ if }n\text{ is not a multiple of 8} \\-1,&\text{ if }n\text{ is a multiple of 8},\end{cases},$$ so that your series is $\sum_{n=1}^\infty \frac{a_n}{n}.$

Then for any natural number $N,$

\begin{align} \sum_{n=1}^{8N}\frac{a_n}{n} &= \sum_{n=1}^{8N}\frac1{n}-2\sum_{n=1}^N \frac1{8n} \\&=\sum_{n=1}^{8N}\frac1{n}-\frac1{4}\sum_{n=1}^N \frac1{n} \\&\ge\sum_{n=1}^{8N}\frac1{n}-\frac1{4}\sum_{n=1}^{8N} \frac1{n} \\&\quad\scriptsize{\quad\text{(because we can only be subtracting *more* positive numbers)}} \\&=\frac3{4}\sum_{n=1}^{8N}\frac1{n}, \end{align}

which approaches $\infty$ as $N$ approaches $\infty,$ since the harmonic series diverges.

  • 14
    $\begingroup$ Thank you, I learned something very important; you can manipulate even conditionally convergent series as long as you manipulate them in the context of partial sums. $\endgroup$ – Ovi Oct 21 '16 at 4:52
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    $\begingroup$ @Ovi -- Yes, because the partial sums are just regular finite sums. But you still need to be careful that your partial sums are entire initial parts of the infinite series. You can't pick and choose which terms to include; all you can do is chop the infinite series off at some finite point. After that, you can apply any normal algebraic operations, because you just have a finite sum. $\endgroup$ – Mitchell Spector Oct 21 '16 at 4:56
  • $\begingroup$ @ Mitchell Spectator Okay got it $\endgroup$ – Ovi Oct 21 '16 at 4:57
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    $\begingroup$ And if you want to prove that the conditionally convergent series does converge, you can't cherry-pick which of the partial sums you're looking at (like it happens here), of course. $\endgroup$ – hmakholm left over Monica Oct 21 '16 at 14:21
  • $\begingroup$ @HenningMakholm Good point. $\endgroup$ – Mitchell Spector Oct 21 '16 at 15:37

To expand on Mitchell Spector's answer, let's generalize to the problem where every $b$th term is negated.

By the same argument, we get $$\begin{align} \sum_{n=1}^{bN}\frac{a_n}{n} &= \sum_{n=1}^{bN}\frac1{n}-2\sum_{n=1}^N \frac1{bn} \\&=\sum_{n=1}^{bN}\frac1{n}-\frac2{b}\sum_{n=1}^N \frac1{n} \\&\ge\sum_{n=1}^{bN}\frac1{n}-\frac2{b}\sum_{n=1}^{bN} \frac1{n} \\&=\left(1-\frac2{b}\right)\sum_{n=1}^{bN}\frac1{n} \end{align}$$

This is nice, since we can see that the divergence proof holds for $b \geq 3$, but fails for $b=2$. This is not strange, since the series for $b=2$ is known to converge to $\log 2$.


This proof isn't valid because of the Riemann series theorem; the original series could still converge even though some rearrangement of it diverges. It's important to focus on the partial sums.

You can argue along the lines that the series $$\sum_{k=0}^{\infty} \frac{1}{8k+1} = 1 + \frac{1}{9} + \frac{1}{17} + \cdots$$ already diverges by comparing its partial sums to the partial sums of the harmonic series $\frac{1}{8} \Big( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \Big).$


Note that $$ \begin{align} &{\tiny\sum_{k=0}^\infty}{\tiny\left(\frac1{8k+1}+\frac1{8k+2}+\frac1{8k+3}+\frac1{8k+4}+\frac1{8k+5}+\frac1{8k+6}+\frac1{8k+7}-\frac1{8k+8}\right)}\\ &={\tiny\sum_{k=0}^\infty}{\tiny\left(\frac1{8k+1}+\frac1{8k+2}+\frac1{8k+3}+\frac1{8k+4}+\frac1{8k+5}+\frac1{8k+6}\right)+{\tiny\sum_{k=0}^\infty}\left(\frac1{8k+7}-\frac1{8k+8}\right)}\\ \end{align} $$ For the left hand sum, we have $$ \begin{align} &{\small\sum_{k=0}^\infty\left(\frac1{8k+1}+\frac1{8k+2}+\frac1{8k+3}+\frac1{8k+4}+\frac1{8k+5}+\frac1{8k+6}\right)}\\ &\ge{\small\frac68\sum_{k=0}^\infty\frac1{k+1}} \end{align} $$ which diverges.

For the right hand sum, we have $$ \begin{align} &{\small\sum_{k=0}^\infty\left(\frac1{8k+7}-\frac1{8k+8}\right)}\\ &\le{\small\frac1{56}+\frac18\sum_{k=1}^\infty\left(\frac1{8k}-\frac1{8k+8}\right)}\\ &=\frac{15}{448} \end{align} $$ A divergent sum plus a convergent sum diverges.

  • $\begingroup$ How did you evaluate $\dfrac 18 \sum_{k=1}^{\infty} \left( \dfrac {1}{8k}-\dfrac {1}{8k+8} \right) = \dfrac {1}{64} \sum_{k=1}^{\infty} \dfrac {1}{k(k+1)}$? I guess its not necessary to calculate the sum because we know it converges anyway, but I'm curious how you got a rational answer, knowing that $\sum_{k=1}^{\infty} \dfrac {1}{k^2} = \dfrac {\pi^2}{6}$ $\endgroup$ – Ovi Oct 24 '16 at 15:19
  • $\begingroup$ $$ \begin{align} \sum_{k=1}^\infty\frac1{k(k+1)} &=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)\\ &=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1k-\sum_{k=2}^{n+1}\frac1k\right)\\ &=\lim_{n\to\infty}\left(1-\frac1{n+1}\right)\\ &=1 \end{align} $$ which is a Telescoping Sum. $\endgroup$ – robjohn Oct 24 '16 at 15:50

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