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Math Help - math homework help: infinite series

  1. #1
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    math homework help: infinite series

    Hello!

    I've been doing math homework and am going over the problems I had difficulty completing. I'm on infinite series. I'd greatly appreciate a walkthrough/explanation for the following problem and some tips on how to approach problems like this/proof problems in general.

    Evaluate: **"<=" means "less than or equal to"; "An" means "Ace of n"

    lim ((2^n)+(3^n))^(1/n) Hint: Show that 3<=An<=(2 x 3^n)^(1/n)
    n-->infinite


    This is an even problem, so I do not have a solution. I tried solving it the easy way, by which I mean setting n equal to infinite, making the equation inside to the zero power, which in turn makes the limit 1. However, because I did not use the "hint," I fear that I may have done something wrong.

    Thank you!
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  2. #2
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    Quote Originally Posted by wyhwang7 View Post
    Hello!

    I've been doing math homework and am going over the problems I had difficulty completing. I'm on infinite series. I'd greatly appreciate a walkthrough/explanation for the following problem and some tips on how to approach problems like this/proof problems in general.

    Evaluate: **"<=" means "less than or equal to"; "An" means "Ace of n"

    lim ((2^n)+(3^n))^(1/n) Hint: Show that 3<=An<=(2 x 3^n)^(1/n)
    n-->infinite
    This is a sequence problem not a series problem!

    3^n \leq 2^n + 3^n \leq 3^n + 3^n = 2\cdot 3^n \implies 3 \leq \left( 2^n + 3^n\right)^{1/n} \leq 3 \cdot 2^{1/n}
    Now use the squeeze theorem.

    This problem has a simple generalization. Let a_1,...,a_k be positive real numbers.
    Then (a_1^n+...+a_k^n)^{1/n} \to \max\{ a_1,...,a_k\}. Try to see if you can prove it.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    This is a sequence problem not a series problem!

    3^n \leq 2^n + 3^n \leq 3^n + 3^n = 2\cdot 3^n \implies 3 \leq \left( 2^n + 3^n\right)^{1/n} \leq 3 \cdot 2^{1/n}
    Now use the squeeze theorem.

    This problem has a simple generalization. Let a_1,...,a_k be positive real numbers.
    Then (a_1^n+...+a_k^n)^{1/n} \to \max\{ a_1,...,a_k\}. Try to see if you can prove it.
    Thank you for your help! I see how you arrived at
    3 <= (2^n + 3^n)^(1/n) <= (2 x 3^n)^(1/n) by raising everything to the power of 1/n.

    Using the Squeeze Theorem, would that make the limit 3?


    Oh, and how did you know to use 3^n <= 2^n + 3^2 <= 3^n + 3^n

    That is, how did you derive the values of Bn and Cn?
    Last edited by mr fantastic; February 19th 2009 at 02:42 AM. Reason: Merged posts
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  4. #4
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    Quote Originally Posted by wyhwang7 View Post
    Thank you for your help! I see how you arrived at
    3 <= (2^n + 3^n)^(1/n) <= (2 x 3^n)^(1/n) by raising everything to the power of 1/n.

    Using the Squeeze Theorem, would that make the limit 3?
    Yes . This is because 2^{1/n}\to 1.

    Oh, and how did you know to use 3^n <= 2^n + 3^2 <= 3^n + 3^n

    That is, how did you derive the values of Bn and Cn?
    I just thought of what I can use to simplify the expression 2^n+3^n. And I saw that if you forget about the 2^n then the expression because 3^n which is easier to work with than 2^n+3^n. Likewise when I wrote 2^n+3^n\leq 2\cdot 3^n, again it made the expression easier to work with being under 1/n power.
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