Question 3 from Exercises 2.5.1 from F. Mary Hart - Guide to Analysis The question tells you that $(a_{3n})^{\infty}_{n=1}$ $(a_{3n+1})^{\infty}_{n=0}$ $(a_{3n+2})^{\infty}_{n=0}$ all converge to a. The question asks you to prove $(a_n)^{\infty}_{n=1}$ converges.  Intuitively this makes sense because the sub-sequence of every 3rd term converges. I know how to get from a sequence converging to its sub-sequence converging but I am not sure about the converse.
 A: *

*Let $\varepsilon>0$ be given.


*Apply the definition of a sequence converging to $a$ for each subsequence:

*

*There is a positive integer $N_1$ so that $|a_{3n}-a|<\varepsilon$ whenever $n \geq N_1$.


*There is a positive integer $N_2$ so that $|a_{3n+1}-a|<\varepsilon$ whenever $n \geq N_2$.


*There is a positive integer $N_3$ so that $|a_{3n+2}-a|<\varepsilon$ whenever $n \geq N_3$.




*Find a formula for a positive integer $N$ that involves the positive integers in step 2 so that the desired conclusion will follow:
$$|a_k-a|<\varepsilon \text{ whenever } k \geq N$$

[Extra Hint]  In determining a formula for $N$, consider restating what we obtain from each of the three hypotheses in step 2 as done for the first of the three below:
In other words, if for $n \in \mathbb{N}$ we set $k_n=3n,$ then $|a_{k_n}-a|<\varepsilon$ whenever $k_n \geq k_1(N_1)=3N_1.$
Also keep in mind that every positive integer $k$ will be equal to the index $k_n$ of some term $a_{k_n}$ in either one of the three subsequences.
