Decide whether this is true: if the series of grouped terms (a_{3n-2} + a_{3n-1} + a_{3n}) converges and a_n tends to 0, then the series of a_n converges. Prove it or give a counterexample.

Step-by-step solution

1. 1

   Translate both hypotheses into statements about partial sums. Let $S_N=\sum_{k=1}^{N}a_k$. Grouping the terms three at a time does not reorder or omit anything, so the $n$-th partial sum of the grouped series is exactly

   $\left(a_1+a_2+a_3\right)+\cdots+\left(a_{3n-2}+a_{3n-1}+a_{3n}\right)=S_{3n}$

   So the first hypothesis says precisely that the subsequence $S_{3n}$ converges, to some limit $S$.

2. 2

   Recognise what is still missing. A subsequence converging does not by itself make the full sequence converge — the other partial sums $S_{3n+1}$ and $S_{3n+2}$ must be shown to approach the same limit. That is exactly what the second hypothesis is for.

3. 3

   Handle the two intermediate partial sums.

   $S_{3n+1}=S_{3n}+a_{3n+1},\qquad S_{3n+2}=S_{3n}+a_{3n+1}+a_{3n+2}$

   Since $a_k\to 0$, both $a_{3n+1}\to 0$ and $a_{3n+2}\to 0$, so

   $S_{3n+1}\to S+0=S,\qquad S_{3n+2}\to S+0+0=S$

4. 4

   Combine the three subsequences. Every index $N$ is of the form $3n$, $3n+1$ or $3n+2$, and all three subsequences of $\{S_N\}$ tend to the same limit $S$. A sequence whose every index falls into finitely many subsequences all converging to $S$ itself converges to $S$:

   $\lim_{N\to\infty}S_N=S\quad\Longrightarrow\quad \sum_{n=1}^{\infty}a_n=S$

   $\boxed{\text{The statement is TRUE}}$

5. 5

   Note why the second hypothesis cannot be dropped. Without $a_n\to 0$ the claim fails: take $a_n=1,-1,0,1,-1,0,\ldots$ — each group of three sums to $0$ so the grouped series converges, but the partial sums $S_N$ cycle through $1,0,0,1,0,0,\ldots$ and never settle. Here $a_n\not\to 0$, which is exactly the escape route the hypothesis closes.

6. 6

   Beware a commonly published "counterexample". Some worked solutions offer $a_n=\tfrac{(-1)^n}{n}$ as a counterexample, but $\sum\tfrac{(-1)^n}{n}$ converges (to $-\ln 2$) by the alternating series test, so it disproves nothing. Any genuine counterexample would have to violate one of the two hypotheses — and by the proof above, none exists.