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Math Help - Convergent Sequence

  1. #1
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    Convergent Sequence

    Suppose a,b>0. Does the sequence {(a^n + b^n)^(1/n)} from n=1 to infinity converge? If so, find the limit.

    Not sure where to start...
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  2. #2
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    Re: Convergent Sequence

    Factor out the larger of a and b. You'll need to consider the two cases where they're equal and not equal separately.

    If the one case isn't clear to you, you could use the identity p = e^{\ln(p)} \ \forall \ p>0, and then look at the limit in the exponent.

    Question:

    \text{If } \lim_{n \to \infty} q_n \text{ exists and equals } L,  \text{ then } \lim_{n \to \infty} e^{q_n}  = e^L \text{ is true because }

    \text{ the function } f(x) = e^x \text{ is ?what? }
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  3. #3
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    Re: Convergent Sequence

    Quote Originally Posted by lovesmath View Post
    Suppose a,b>0. Does the sequence {(a^n + b^n)^(1/n)} from n=1 to infinity converge? If so, find the limit.
    This is such a well known problem.
    Suppose that a>b>0 then a = \sqrt[n]{{a^n }} \leqslant \sqrt[n]{{a^n  + b^n }} \leqslant a\sqrt[n]{2} \to a

    It can be extended to more than two variables.
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    Re: Convergent Sequence

    How did you get the 2 under the radical at the end of the inequality?
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    Re: Convergent Sequence

    Quote Originally Posted by lovesmath View Post
    How did you get the 2 under the radical at the end of the inequality?
    It is simple arithmetic.
    Because b^n\le a^n we have \sqrt[n]{{a^n  + b^n }} \leqslant \sqrt[n]{{a^n  + a^n }} = \sqrt[n]{{2a^n }} = a\sqrt[n]{2}
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