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Prove that sequence converges

  1. Nov 16, 2011 #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.
     
  2. jcsd
  3. Nov 16, 2011 #2

    Dick

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    Start by factoring 3^n out of the radical. Use some algebra.
     
  4. Nov 16, 2011 #3
    Can you get any results. Is the sequence bounded? Is it monotonically increasing or decreasing? etc?
     
  5. Nov 16, 2011 #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??
     
  6. Nov 16, 2011 #5

    Dick

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    Algebra. Factor 2^n+3^n=(3^n)*(something). What's the (something)? Find out by doing algebra.
     
  7. Nov 16, 2011 #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?
     
  8. Nov 16, 2011 #7
    I think I figured it out. would it be 3 * nth root of ((3/2)^n+1)?
     
  9. Nov 16, 2011 #8

    Dick

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    Close. Can you show how you got there? Why did you get (3/2)^n instead of (2/3)^n?
     
  10. Nov 16, 2011 #9
    oops i meant to write that. so from here do i need to find the N?
     
  11. Nov 16, 2011 #10

    Dick

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    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?
     
  12. Nov 16, 2011 #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
     
  13. Nov 16, 2011 #12

    Dick

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    So? What would that make the limit? What's nth root of 1?
     
  14. Nov 16, 2011 #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
     
  15. Nov 16, 2011 #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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