# Finding $\lim_{n \rightarrow \infty}a_n$ given $\lim_{n \rightarrow \infty}\frac{a_n -1}{a_n + 1}$

Let $$(a_n)_{n=1}^{\infty}$$ be a real sequence. I am given $$\lim_{n \rightarrow \infty}\frac{a_n-1}{a_n + 1} = 0$$ (note that this means we cannot have $$a_n = -1$$), and I want to accordingly show that $$\lim_{n \rightarrow \infty}a_n = 1$$. Here is my attempt:

Let $$b_n = \frac{a_n-1}{a_n + 1}$$ $$\forall n \in \mathbb N$$. This gives $$a_n = \frac{1+b_n}{1-b_n}$$ $$\forall n \in \mathbb N$$. Now this expression for $$a_n$$ is well-defined as $$\nexists n \in \mathbb N$$ s.t. $$b_n = 1$$; if such an $$n$$ existed we would have $$a_n = \frac{2}{0}$$, which contradicts that $$a_n$$ is a real sequence.

Thus we have that $$a_n = \frac{1+b_n}{1-b_n} = \frac{1+0}{1-0} = 1$$ as $$n \rightarrow \infty$$. Thus we have $$\lim_{n \rightarrow \infty}a_n = 1$$.

Now I am told that my argument above is invalid because it's possible that $$b_n = 1$$, and that the argument below is actually the correct argument:

Let $$b_n = \frac{a_n-1}{a_n + 1}$$ $$\forall n \in \mathbb N$$. As $$b_n\rightarrow 0$$ as $$n\rightarrow\infty$$, taking $$\varepsilon=1$$ in the definition of convergence, there exists $$n_0\in\mathbb N$$ such that for $$n\in\mathbb N$$ with $$n\geq n_0$$, $$|b_n|<1$$; in particular for $$n\geq n_0$$, $$b_n\neq 1$$. Thus we have $$a_n = \frac{1+b_n}{1-b_n}$$ $$\forall n \in \mathbb N$$ s.t. $$n \geq n_0$$. This then gives $$a_n = \frac{1+b_n}{1-b_n} = \frac{1+0}{1-0} = 1$$ as $$n \rightarrow \infty$$. Thus we have $$\lim_{n \rightarrow \infty}a_n = 1$$.

Now, I perfectly understand the second argument but I can't understand why first argument is wrong as I have proved, by contradiction, that there cannot be $$b_n = 1$$.

Let $$\lim\limits_{n\rightarrow+\infty}\frac{a_n-1}{a_n+1}=a.$$
Thus, for $$a\neq1$$ we obtain: $$\lim_{n\rightarrow+\infty}a_n=\lim_{n\rightarrow+\infty}\frac{1+\frac{a_n-1}{a_n+1}}{1-\frac{a_n-1}{a_n+1}}=\frac{1+a}{1-a}.$$
• I was aware of this, but I'm more concerned whether in your argument it is necessary to first restrict $n$ to greater than or equal to $n_0$, where for $n > n_0$ we have $\frac{a_n-1}{a_n+1} \neq 1$, or whether we implicitly cannot have, for all $n$, $\frac{a_n-1}{a_n+1} = 1$ in the first place. – Hai Feb 9 at 10:12
• @Hai Yes for $n>n_0$ it happens by the definition of the limit. If you want, you can write it, but for me it's not necessary. – Michael Rozenberg Feb 9 at 10:18
• @Thanks, but I was also wondering whether the condition of $n > n_0$ is necessary because from my working I have found that we implicitly have $\frac{a_n-1}{a_n+1} \neq 1$ for all $n$. – Hai Feb 9 at 10:57
• @Hai If $a=1$ then easy to see that $\lim\limits_{n\rightarrow+\infty}a_n$ does not exist. – Michael Rozenberg Feb 9 at 11:00