# Convergence Proof need help

1. Oct 23, 2014

### lmao2plates

• OP has been warned that posts need to show what was attempted
1. The problem statement, all variables and given/known data

an = [sin(n)]/n

Prove that this sequence converges using Cauchy theorem

2. Relevant equations

Cauchy theorem states that:

A sequence is called a Cauchy theorem if for all ε > 0, there exists N , for all n > N s.t. |xn+1 - xn| < ε

I do not know how to approach this proof.

I would appreciate some help.

2. Oct 23, 2014

### RUber

Can you think of an inequality that would apply to $\left|\frac {\sin(n)}{n}-\frac {\sin(n+1)}{n+1}\right|$?

3. Oct 23, 2014

### RUber

You should end up with a statement that says, given $\epsilon>0$
$\left|\frac {\sin(n)}{n}-\frac {\sin(n+1)}{n+1}\right| < \epsilon$ for any $n \geq N \geq f(\epsilon)$

4. Oct 23, 2014

### pasmith

That is not the correct definition. You have stated the condition for $x_{n+1} - x_n \to 0$, which is a necessary but not a sufficient condition for $x_n$ to converge.

The correct definition is:

A sequence $(a_n)$ is Cauchy if and only if for every $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that for all $n \in \mathbb{N}$ and all $m \in \mathbb{N}$, if $n \geq N$ and $m \geq N$ then $|a_n - a_m| < \epsilon$.

There is a theorem which states that a real sequence converges if and only if it is Cauchy.

You may find it helpful show that $|m \sin n - n \sin m| < n + m$.

5. Oct 23, 2014

### GFauxPas

I used $0 \le |\sin x| \le 1$