Proving Convergence of an = [sin(n)]/n w/ Cauchy Theorem

[lmao2plates]

Homework Statement

a_n = [sin(n)]/n

Prove that this sequence converges using Cauchy theorem

Homework 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. |x_n+1 - x_n| < εI do not know how to approach this proof.

I would appreciate some help.

 

[RUber]

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

 

[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 $n \geq N \geq f(\epsilon)$

 

[pasmith]

Homework Statement

a_n = [sin(n)]/n

Prove that this sequence converges using Cauchy theorem

Homework 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. |x_n+1 - x_n| < ε

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 &gt; 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| &lt; \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| &lt; n + m.

 

[GFauxPas]

You may find it helpful show that |m \sin n - n \sin m| &lt; n + m.

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