Trying to prove that the sequence: 3, 3, sin(1), 3, 3, sin(2), 3, 3, sin(3), 3, 3, sin(4), $\ldots$ does not converge to 3

Question

$\textbf{Proof}$:

I need to show that $\exists \epsilon > 0$, such that, $\forall N \in \mathbb{N},\ \exists n \ge N$, such that, $\ \left | a_n - 3 \right | \ge \epsilon$

Let $\epsilon = 2$, then if you take $n = 30$, we have that $a_{30} = sin(10) \approx -0.544 \ldots$

So we have that $ \left | sin(10) - 3 \right | \ge 2$

Therefore I have showed that, $\exists \epsilon > 0$, such that, $\forall N \in \mathbb{N},\ \exists n \ge N$, such that, $\ \left | a_n - 3 \right | \ge \epsilon$, where $\epsilon = 2$ and $n = 30$.

I am wondering if this proof is right.

Answer 1

You have to find $\varepsilon>0$ such that, for all $N\in\mathbb N$, there exists $n\geq N$ such that $|a_n-3|\geq \varepsilon$. What you did will only work for $N\leq 30$, but, if for example $N=31$, you have to find an $n\geq 31$ such that $|a_n-3|\geq\varepsilon$.

Instead, you could do the following: given $N\in\mathbb N$, there exists $n\geq N$ such that $a_n=\sin k$ for some $k\in\mathbb N$. So, for that $n$, $|a_n-3|=|\sin k-3|\geq\left||3|-|\sin k|\right|\geq 3-|\sin k|\geq 2$.

Answer 2

It suffices you show $\langle \sin n\rangle $ cannot converge to $3$. That is, if the sequence converged to $3$, then the subsequence $\sin 1,\sin 2,\ldots$ would converge to $3$. But for any $n$, we have that $\sin n \leqslant 1$. Thus, this is impossible. Note that if we replaced $3$ with $1$; things'd get much more delicate!