No convergent sequence containing infinitely many ones can have a limit other than one.

Question

Is the following argument correct?

Proposition. There cannot exist a sequence with an infinite number of ones that converges to a limit not equal to one.

Proof. Let $(a_n)$ be a sequence containing an infinite number of ones such that $(a_n)\to L$ where $L\neq 1$, then $|1-L|>0$ and from hypothesis there exists an $N\in\mathbf{N}$ such that $|a_n-L|<|1-L|/2,\forall n\ge N$.

In addition there must exist a $k\in\{N,N+1,N+2,\dots\}$ such that $a_{k}=1$, otherwise the sequence may only have finitely many ones, but then $|a_k-L| = |1-L|<|1-L|/2$ a contradiction.

$\blacksquare$

Answer 1

The proof is correct. You can generalize it as follows:

Suppose the sequence $(a_n)$ is convergent to $L$. If $b\ne L$, then the set $\{n\in\mathbb{N}:a_n=b\}$ is finite.

Proof. Suppose $b<L$ and set $\varepsilon=(L-b)/2$. Then there exists $N$ such that $|a_n-L|<\varepsilon$, for every $n\ge N$. This implies $$ -\frac{L-b}{2}+L<a_n<\frac{L-b}{2}+L $$ in particular $a_n>(L+b)/2>b$, for every $n\ge N$.

Similarly for $b>L$. QED