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: $$\lim_{n\rightarrow+\infty}a_n=\lim_{n\rightarrow+\infty}\frac{1+\frac{a_n-1}{a_n+1}}{1-\frac{a_n-1}{a_n+1}}=\frac{1+a}{1-a}.$$

  • $\begingroup$ 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. $\endgroup$ – Hai Feb 9 at 10:12
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    $\begingroup$ @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. $\endgroup$ – Michael Rozenberg Feb 9 at 10:18
  • $\begingroup$ @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 $n$. $\endgroup$ – Hai Feb 9 at 10:57
  • $\begingroup$ @Hai If $a=1$ then easy to see that $\lim\limits_{n\rightarrow+\infty}a_n$ does not exist. $\endgroup$ – Michael Rozenberg Feb 9 at 11:00
  • $\begingroup$ I see, thanks. That was my first impression. $\endgroup$ – Hai Feb 9 at 11:02

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