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 $b_n = 1$.
Best Answer
Let $\lim\limits_{n\rightarrow+\infty}\frac{a_n-1}{a_n+1}=a.$
Thus, for $a\neq1$ we obtain: $$\lim_{n\rightarrow+\infty}a_n=\lim_{n\rightarrow+\infty}\frac{1+\frac{a_n-1}{a_n+1}}{1-\frac{a_n-1}{a_n+1}}=\frac{1+a}{1-a}.$$