# Proof that (n+1)/(3n-1) converges to 1/3

1. May 3, 2010

### kolley

1. The problem statement, all variables and given/known data

prove (n+1)/(3n+1) converges to 1/3

2. Relevant equations

3. The attempt at a solution

I have been trying to figure this out for a while. I started out

(n+1)/3n+1)-1/3=(2n+2)/(9n-3) now I don't know how to proceed with the proof. What do I set N equal to? And then how do work it back into my inequality chain? Working this out step by step would be greatly appreciated.

2. May 3, 2010

Hint: L'Hôpital's rule.

3. May 3, 2010

### kolley

how does L'Hopital's rule get incorporated into a formal proof for a sequence converging. I don't understand

4. May 3, 2010

### HallsofIvy

Are you required to use an "N-$\epsilon$" proof? If not, then you can use L'Hopital's rule- although even then, I would consider it "overkill". If you are not required to use L'Hopital, just divide both numerator and denominator by n and use the fact that $1/n\to 0$ as n goes to infinity.

If you are required to use "N-$\epsilon$":

To start with you are completly wrong in saying that (n+1)/3n+1)-1/3=(2n+2)/(9n-3). I presume the denominator, 9n-3 instead of 9n+3, is a typos but even so,
$$\frac{n+1}{3n+1}- \frac{1}{3}= \frac{3(n+ 1)- (3n+1)(1)}{3(3n+ 1)}= \frac{3n+ 3-3n-1}{9n+ 3}= \frac{2}{9n+1}$$

That is, you need to make
$$\left|\frac{n+1}{3n+1}- \frac{1}{3}\right|= \frac{2}{9n+1}< \epsilon$$

it should be easy to continue from there.