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