# Prove that sequence converges

1. Nov 16, 2011

### atthebeach

If an = $\sqrt[n]{2^n+3^n}$ does the sequence converge? Prove your assertion.

I have no idea where to start with this problem. It does have something to with $\exists$ N such that n>N $\Rightarrow$ |an - a| < ϵ for all ϵ>0

Can yall help me? It would be greatly appreciated.

2. Nov 16, 2011

### Dick

Start by factoring 3^n out of the radical. Use some algebra.

3. Nov 16, 2011

### deluks917

Can you get any results. Is the sequence bounded? Is it monotonically increasing or decreasing? etc?

4. Nov 16, 2011

### atthebeach

It is monotone decreasing. Looking at a graph I can deduce that it converges to 3. but I really struggle with the proof...and the algebra sadly.

How would I factor 3^n out of the radical??

5. Nov 16, 2011

### Dick

Algebra. Factor 2^n+3^n=(3^n)*(something). What's the (something)? Find out by doing algebra.

6. Nov 16, 2011

### Dick

Lack of algebra skills is bad news. Let me put the same problem more simply. Suppose the expression were a+b=a*x. What's x?

7. Nov 16, 2011

### atthebeach

I think I figured it out. would it be 3 * nth root of ((3/2)^n+1)?

8. Nov 16, 2011

### Dick

Close. Can you show how you got there? Why did you get (3/2)^n instead of (2/3)^n?

9. Nov 16, 2011

### atthebeach

oops i meant to write that. so from here do i need to find the N?

10. Nov 16, 2011

### Dick

Not necessarily. I don't think this is probably an epsilon type of proof, unless the problem says it is. You just need an argument that says it's true. lim n->inf (2/3)^n=0, right?

11. Nov 16, 2011

### atthebeach

yes. but then i have the 1 under the radical and i know that the infinith root of 1 would be 1. but....i cannot split up the terms under the radical

12. Nov 16, 2011

### Dick

So? What would that make the limit? What's nth root of 1?

13. Nov 16, 2011

### atthebeach

OHHH the limit would be three thank you so much. I think maybe i was supposed to do an epsilon proof but this should be sufficient

14. Nov 16, 2011

### Dick

Very welcome. I think an epsilon type proof would be more challenging than the problem actually calls for.