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My Solution

Let the subscript begin at 1. Group the terms three by three. The partial sum $S_{n}$ satisfies

$$S_{3n}-S_{3(n-1)}=\frac{1}{3n-2}+\frac{1}{3n-1}-\frac{1}{3n} > \frac{1}{3n}+\frac{1}{3n}-\frac{1}{3n}=\frac{1}{3n},\quad(n\ge1),$$

where $S_0=0$. Thus $S_{3n}>\sum\limits_{i=1}^n \frac{1}{3i}$ and $S_{3n}$ diverges as $n\to+\infty$. As $\{S_{3n}\}$ is a divergent subseries (?) of $\{S_n\}$, so the original series must be divergent.


  1. Is the word "subseries" correct? [Edit: Yeah, I find "subsequence" a better word.]
  2. Is my solution a rigorous proof? [Edit: It is a proof. It has no flaw. It is rigorous.]
  3. Are there other different / elegant / interesting solutions? [Edit: Seems that my solution is succinct enough.]
  4. What is a rigorous proof? [Edit: A proof that has no flaw is rigorous.]

Thank you all!

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+1 for showing your thoughts (and good ones). –  Ross Millikan Oct 20 '12 at 4:27
I think "subseries" is fine in this context since there isn't really a chance of confusion. If you really want to be formal, I would define the sequence of partial sums $(S_n)$ and then just call $(S_{3n})$ a subsequence of $(S_n)$. –  EuYu Oct 20 '12 at 4:27
Thank you. Can someone show me how I can apply Cauchy's convergence test? I don't know how to combine the three different cases of $n \mod 3 = 0,1,2$. –  FrenzY DT. Oct 20 '12 at 4:48
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1 Answer

up vote 1 down vote accepted

I think subseries is a fine word, saying exactly what you want. I would accept your solution-especially as it is the one I would offer.

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Thanks. Would you mind sparing some time for the Cauchy's test? My classmate says that Cauchy's can do the trick too, but I dare not go beyond the definition. –  FrenzY DT. Oct 20 '12 at 4:50
What you have done shows that Cauchy's test will not prove convergence. If I give you $\epsilon$ you can show that adding enough terms far out will exceed that. It is the same proof as the harmonic series, with a factor $3$ that you have identified. –  Ross Millikan Oct 20 '12 at 4:58
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