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

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Homework Help Overview

The discussion revolves around the convergence of the sequence defined by \( a_n = \sqrt[n]{2^n + 3^n} \). Participants are exploring the mathematical properties of this sequence, including its boundedness and monotonicity, as well as the necessary steps to prove convergence.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss factoring \( 3^n \) out of the radical and consider the implications of the sequence being monotone decreasing. Questions arise about the algebra involved in the factorization and the nature of limits.

Discussion Status

There is a productive exchange of ideas, with some participants suggesting algebraic manipulations and others questioning the steps taken. While there is no explicit consensus on the proof method, guidance is provided regarding the limit behavior of the sequence.

Contextual Notes

Some participants express uncertainty about their algebra skills and the requirements for a formal proof, including whether an epsilon-delta argument is necessary. The discussion reflects a range of understanding regarding the convergence proof process.

atthebeach
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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.
 
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Start by factoring 3^n out of the radical. Use some algebra.
 
Can you get any results. Is the sequence bounded? Is it monotonically increasing or decreasing? etc?
 
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??
 
atthebeach said:
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.
 
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?
 
I think I figured it out. would it be 3 * nth root of ((3/2)^n+1)?
 
atthebeach said:
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?
 
oops i meant to write that. so from here do i need to find the N?
 
  • #10
atthebeach said:
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?
 
  • #11
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
atthebeach said:
yes. but then i have the 1 under the radical

So? What would that make the limit? What's nth root of 1?
 
  • #13
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
Very welcome. I think an epsilon type proof would be more challenging than the problem actually calls for.
 

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