Convergence of the Sequence An = √(2^n + 3^n)

[atthebeach]

If a_n = $\sqrt[n]{2^n+3^n}$ 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 $\Rightarrow$ |a_n - a| < ϵ for all ϵ>0

Can yall help me? It would be greatly appreciated.

 

[Dick]

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

 

[deluks917]

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

 

[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??

 

[Dick]

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??

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

 

[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?

 

[atthebeach]

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

 

[Dick]

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

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

 

[atthebeach]

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

 

[Dick]

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

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?

 

[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

 

[Dick]

yes. but then i have the 1 under the radical

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

 

[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

 

[Dick]

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