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Math Help - A question on sequences

  1. #1
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    a_n= (3^n + 5^n)^{\frac 1n}

    is the above sequence convergent? if so, what is the limit?
    Last edited by Jhevon; March 14th 2009 at 10:35 PM.
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    Quote Originally Posted by twilightstr View Post
    a_n= (3^n + 5^n)^{\frac 1n}

    is the above sequence convergent? if so, what is the limit?
    it is convergent.

    Hint: (3^n + 5^n)^{\frac 1n} = \text{exp} \left( \ln (3^n + 5^n)^{\frac 1n} \right) = \text{exp} \left( \frac {\ln (3^n + 5^n)}n \right)

    now take the limit
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  3. #3
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    is it infinity
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    Quote Originally Posted by twilightstr View Post
    is it infinity
    no (if it were, it wouldn't be convergent)

    Hint 2: you can use L'Hopital's rule to find the limit
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  5. #5
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    Just as an alternative..

    If you can use the fact that \lim_{n \to \infty} a^{\frac{1}{n}} = 1 for a > 0, then notice that:

    5^n \ \ \leq \ \ 3^n + 5^n \ \ \leq \ \ 5^n + 5^n

    \Rightarrow \left(5^n \right)^{\frac{1}{n}} \ \ \leq \ \ \left(3^n +5^n \right)^{\frac{1}{n}} \ \ \leq \ \ \left(5^n +5^n \right)^{\frac{1}{n}}

    \Rightarrow 5  \ \ \leq \ \ \left(3^n +5^n \right)^{\frac{1}{n}} \ \ \leq \ \ 5\left(2 \right)^{\frac{1}{n}}

    And use squeeze theorem ..
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  6. #6
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    jhevon, how would u take the limit
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    Quote Originally Posted by twilightstr View Post
    jhevon, how would u take the limit
    I told you. I would use L'Hopital's rule. o_O gives a nice alternative
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    yes. thats what i tried doing in the first place, but i think i did it incorrectly.
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    Quote Originally Posted by twilightstr View Post
    yes. thats what i tried doing in the first place, but i think i did it incorrectly.
    recall that, by L'Hopital's, \lim_{n \to \infty} \frac {\ln (3^n + 5^n)}n = \lim_{n \to \infty} \frac {\frac d{dn}[ \ln (3^n + 5^n)]}{\frac d{dn}n}

    i suppose it is the \frac d{dn}[ \ln (3^n + 5^n)] that is giving you trouble. what did you get for this?
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  10. #10
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    Talking

    Quote Originally Posted by twilightstr View Post
    yes. thats what i tried doing in the first place, but i think i did it incorrectly.
    We'll be glad to look for any errors, but you'll need to show the work you did.

    Please be complete. Thank you!
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  11. #11
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    = [(ln3)3^x + (ln5)5^x]/(3^x + 5^x) as the limit goes to infinity. I really dont know what to do from there
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    Quote Originally Posted by twilightstr View Post
    = [(ln3)3^x + (ln5)5^x]/(3^x + 5^x) as the limit goes to infinity. I really dont know what to do from there
    well, after that, we're good. multiply by \frac {\frac 1{5^x}}{\frac 1{5^x}}, the limit should seem obvious from there
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  13. #13
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    \left( 3^{n}+5^{n} \right)^{1/n}=5\left\{ \left( \frac{3}{5} \right)^{n}+1 \right\}^{1/n}\to 5 since \left( \frac{3}{5} \right)^{n}+1\to 1 and 1^{1/n}\to 1 as n\to\infty.
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