Does $a_n$ converge, and, if so, what is the limit? Let, for n $\in \mathbb{N}$
$$a_n = \begin{cases}
        \frac{1}{n} & \text{if n = 10 $\mathbb{m}$, for $\mathbb{m} \in \mathbb{N}$ }
        \\ 
        \frac{sin(n)}{n} & \text{otherwise}
        \end{cases}.
$$
Does($a_n$) converge, and if so, what is the limit?
So the limit of $\frac{1}{n}$ as n $\to \infty$ is 0 and that is the same for $\frac{sin(n)}{n}$.
I believe that since $\frac{1}{n}$ only appear every so often and that it also converge to 0, the limit of the sequence of is 0. Is my reasoning rigorous enough?
 A: No, it is not rigorous enough, because “only appear every so often” is not rigorously defined. (Besides, if it appeared half the times, the limit would also be $0$; and so, it is not really important if it appears rarely or very often.) But you can say that, for each $n\in\Bbb N$, $0\leqslant|a_n|\leqslant\frac1n$, and that threfore, by the squeeze theorem, $\lim_{n\to\infty}|a_n|=0$. And this is the same thing as claiming that $\lim_{n\to\infty}a_n=0$.
A: You write that $1/n$ "only appears every so often", which is true. But this observation is too weak to be considered a rigorous justification for your conclusion, without saying more. (In fact, I don't think it even gives a convincing non-rigorous argument.)
Nevertheless your observation provides a starting point for something rigorous and convincing. Let $x_n = 1/n$ and $y_n=\sin n / n$. As you correctly note, these two are null sequences (although see my remarks below). Now form the sequences $s_n$:
$$0,\;0,\;0,\;0,\;0,\;0,\;0,\;0,\;0,\;\frac1{10},\\ \;\;\;\;\;\;\;\;\;\;0,\;0,\;0,\;0,\;0,\;0,\;0,\;0,\;0,\;\frac1{20},\; 0,\; \dots .$$
and $t_n:$
$$\frac{\sin 1}1,\;\frac{\sin 2}2,\;\frac{\sin 3}3,\;\frac{\sin 4}4,\;\frac{\sin 5}5,\;\frac{\sin 6}6,\;\frac{\sin 7}7,\;\frac{\sin 8}8,\;\frac{\sin 9}9,\;0,\\ \;\;\;\;\;\;\;\;\;\;\frac{\sin 11}{11},\dots .$$
This makes more explicit your notion of only appearing every so often, which applies to both $x_n$ and $y_n$ as they interleave to form $a_n$. Note that $a_n = s_n + t_n.$ It is well known that

*

*The sum of two null sequences is again a null sequence.  $\quad$ (*)

So we are done if we can show that each summand is a null sequence (i.e. that each converges to zero). But this follows from a second useful fact:


*The product of a bounded sequence with a null sequence is again a null sequence.  $\quad$ (**)

We see that $s_n = b_n x_n$, where $b_n$ is a suitable sequence of zeros and ones, "turning on" $x_n$ when it is due to "appear". Similarly, $t_n = B_n y_n$ for suitable bounded $B_n$.
Remarks. You have taken for granted that $\sin n / n$ is a null sequence. The usual way to show this is via $-1 \leqslant \sin x \leqslant 1$ for any real $x$, i.e. $|\sin n| \leqslant 1$ invariably. Another answer already posted shows that in exactly the same way, we can directly solve your problem, without any appeal to (*) and (**). However, I wished to show a rigorous argument that makes use of your observation about the appearing/disappearing or "taking turns" structure of $a_n$.
A: No, your reasoning is not rigorous enough and the argument of the frequency of $\frac 1n$ is not pertinent here since convergence deals with infinitely many terms and $\frac 1n$'s will appear infinitely many times. For your reasoning to be perfectly rigorous, you can either prove very formally using the definition of convergence that $$\forall \epsilon > 0, \exists N \in N, \forall n \geq N, \quad \lvert a_n \rvert \leq \epsilon $$  Or you can just notice that $0 \leq \lvert a_n \rvert\leq \frac 1n$ $\forall n \neq 0$ and use squeeze theorem.
