# Test the convergence of: $1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots$

### 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.

### Questions

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

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