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Does the series $$S=\left(1+\frac{1}{2}-\frac{1}{3} \right) + \left(\frac{1}{4}+\frac{1}{5}-\frac{1}{6} \right)+\left(\frac{1}{7}+\frac{1}{8}-\frac{1}{9}\right)+\cdots$$ converge?

Here's my attempt at a solution: $$S = \sum_{n=1}^{\infty}\frac{1}{n}-2\sum_{n=1}^{\infty}\frac{1}{3n}=\sum_{n=1}^{\infty}\frac{1}{3n}=\frac{1}{3}\sum_{n=1}^{\infty}\frac{1}{n}$$

As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of $S$.

Is this right? Which other convergence tests could be used?

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    $\begingroup$ your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series $\endgroup$
    – Masacroso
    Jan 6, 2019 at 3:15
  • $\begingroup$ @Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge". $\endgroup$
    – John
    Jan 6, 2019 at 3:21
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    $\begingroup$ @Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round). $\endgroup$
    – AlephNull
    Jan 6, 2019 at 12:29
  • $\begingroup$ Of course, you're correct. For whatever reason, I thought each term was positive.. $\endgroup$
    – John
    Jan 7, 2019 at 15:19

2 Answers 2

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Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows: $$\sum_{n=0}^\infty\left(\frac{1}{3n+1}+\frac{1}{3n+2}-\frac{1}{3n+3}\right)$$ Simplify what is in the parentheses and then evaluate in the usual way.

Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.

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    $\begingroup$ In this case you can simply say $\frac 1{3n+2}-\frac 1{3n+3}>0$ thus the series is $>\sum\frac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $\frac {9n^2}{27n^3}$. $\endgroup$
    – zwim
    Jan 6, 2019 at 4:32
  • $\begingroup$ @zwim quite true, that is a nice little shortcut $\endgroup$
    – Ben W
    Jan 6, 2019 at 14:47
  • $\begingroup$ Simplify and it behaves like $1/n$, so it diverges. $\endgroup$ Jan 6, 2019 at 18:12
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The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.

Let $H_n = \sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the $$ S_n = H_{3n} - \frac 23 H_n, $$ then use the asymptotic expression $H_n = \log n + \gamma + \varepsilon_n$ where $\varepsilon_n \to 0 [n \to \infty]$, we have $$ S_n = \log (3n) -\frac 23\log n + \frac 13 \gamma + \varepsilon_{3n} - \frac 23 \varepsilon_n = \log(3n^{1/3}) + \frac 13 \gamma + \alpha_n \xrightarrow{n \to \infty} +\infty. $$

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