1. ## convergence

If $a_n$ is a sequence such that $\frac{a_{n}-1}{a_{n}+1} \to 0$, then does $\lim_{n \to \infty} a_{n}$ exist? So in other words, does $a_n \to L$?

Assume that $s_n = \frac{a_{n}-1}{a_{n}+1}$. Let $\varepsilon > 0$. Then there exists an $N \in \mathbb{N}$, such that whenever $n \geq N$, then $|s_{n}| < \varepsilon$.

Now the problem is to express $a_{n}$ in terms of $s_{n}$ and use the definition of convergence to see if it converges or not?

2. Hello
Originally Posted by particlejohn
If $a_n$ is a sequence such that $\frac{a_{n}-1}{a_{n}+1} \to 0$, then does $\lim_{n \to \infty} a_{n}$ exist? So in other words, does $a_n \to L$?

Assume that $s_n = \frac{a_{n}-1}{a_{n}+1}$. Let $\varepsilon > 0$. Then there exists an $N \in \mathbb{N}$, such that whenever $n \geq N$, then $|s_{n}| < \varepsilon$.

Now the problem is to express $a_{n}$ in terms of $s_{n}$ and use the definition of convergence to see if it converges or not?
If you have to use the definition of convergence, that's the idea. To help you, $|s_n|<\varepsilon$ can be written $-\varepsilon < 1-\frac{2}{a_n+1} < \varepsilon$.

If you don't have to use the definition of convergence, suppose that $(a_n)$ diverges and see what happens to $s_n=1-\frac{2}{a_n+1}$ which is supposed to tend to 0 as $n\to\infty$.