Prove that sequence converges

  • Thread starter atthebeach
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  • #1
If an = [itex]\sqrt[n]{2^n+3^n}[/itex] does the sequence converge? Prove your assertion.

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

Can yall help me? It would be greatly appreciated.
 

Answers and Replies

  • #2
Dick
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Start by factoring 3^n out of the radical. Use some algebra.
 
  • #3
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4
Can you get any results. Is the sequence bounded? Is it monotonically increasing or decreasing? etc?
 
  • #4
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
Dick
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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.
 
  • #6
Dick
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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
I think I figured it out. would it be 3 * nth root of ((3/2)^n+1)?
 
  • #8
Dick
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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?
 
  • #9
oops i meant to write that. so from here do i need to find the N?
 
  • #10
Dick
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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
Dick
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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
Dick
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Very welcome. I think an epsilon type proof would be more challenging than the problem actually calls for.
 

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