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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.

Ovi
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4 Answers4

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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.

Mitchell Spector
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    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. – Ovi Oct 21 '16 at 04:52
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    @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. – Mitchell Spector Oct 21 '16 at 04:56
  • @ Mitchell Spectator Okay got it – Ovi Oct 21 '16 at 04:57
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    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. – hmakholm left over Monica Oct 21 '16 at 14:21
  • @HenningMakholm Good point. – Mitchell Spector Oct 21 '16 at 15:37
  • @Ovi The definition of an infinite sum is the limit of the partials in the order given. – Simply Beautiful Art Oct 23 '16 at 14:58
  • @HenningMakholm I'm afraid I don't really understand what you mean, could you give an example? I don't see how we cherry picked a partial sum here; we considered the partial sums $s_1 = \dfrac 11, s_2 =\dfrac 11 + \dfrac 12$, $\ldots$. These are the unique partial sums which correspond to $\sum_{n=1}^{\infty} \dfrac 1n$, aren't they? – Ovi Oct 23 '16 at 17:05
  • Not really -- summing "just the even terms" doesn't correspond to any of the partial sums. I'm imagining for example $\sum_{n=0}^\infty (-1)^n$, which obviously doesn't converge -- but if we look only at the partial sums where the number of terms is a multiple of 8, we find that _they_ are all $0$, so we might erroneously conclude that the series converges, @Ovi – hmakholm left over Monica Oct 23 '16 at 17:28
  • @HenningMakholm -- You're right, I've deleted that comment because it didn't address the issue. I had two examples, and mistakenly copied the wrong one. (The right one was the same as yours.) – Mitchell Spector Oct 23 '16 at 17:30
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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 $\log 2$.

Daniel R
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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 $\frac{1}{8} \Big( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \Big).$

user378953
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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.

robjohn
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  • 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 $\sum_{k=1}^{\infty} \dfrac {1}{k^2} = \dfrac {\pi^2}{6}$ – Ovi Oct 24 '16 at 15:19
  • $$ \begin{align} \sum_{k=1}^\infty\frac1{k(k+1)} &=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)\\ &=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1k-\sum_{k=2}^{n+1}\frac1k\right)\\ &=\lim_{n\to\infty}\left(1-\frac1{n+1}\right)\\ &=1 \end{align} $$ which is a [Telescoping Sum](https://en.wikipedia.org/wiki/Telescoping_series). – robjohn Oct 24 '16 at 15:50