Results 1 to 4 of 4

Math Help - math homework help: infinite series

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    25

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2009
    Posts
    25
    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 03:42 AM. Reason: Merged posts
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need Math Homework help?
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: January 13th 2011, 10:19 PM
  2. Math Homework Help
    Posted in the Algebra Forum
    Replies: 7
    Last Post: September 30th 2010, 03:09 PM
  3. Replies: 1
    Last Post: June 2nd 2010, 10:49 AM
  4. Cal II infinite series homework questions...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 2nd 2009, 08:02 PM
  5. Math Homework Help!
    Posted in the Algebra Forum
    Replies: 6
    Last Post: August 21st 2008, 11:38 AM

Search Tags


/mathhelpforum @mathhelpforum