# Thread: convergence

1. ## convergence

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

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

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

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

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

Now the problem is to express $\displaystyle a_{n}$ in terms of $\displaystyle s_{n}$ 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, $\displaystyle |s_n|<\varepsilon$ can be written $\displaystyle -\varepsilon < 1-\frac{2}{a_n+1} < \varepsilon$.

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