Results 1 to 2 of 2

Math Help - Convergent, divergent or definitely divergent?

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    17

    Convergent, divergent or definitely divergent?

    Hi,
    can someone please explain to me how to find out if a set is convergent, divergent, or definitely divergent. I've seen the test to do in a textbook but i don't understand it.

    I'm supposed to find out what these 5 sets are, and find the limit if its convergent.
    I'd appreciate it if you showed me 1 or 2 of them as an example.
    Thank you

    a) an = n^2(n-3)

    b) 2n^3 + 4n^2 - 5 / 8n^3 - 2n

    c) n^2 / n^3 - 4

    d) (-n)^-3

    e) n(-1)^n
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2009
    Posts
    30
    Quote Originally Posted by ihatemath09 View Post
    Hi,
    can someone please explain to me how to find out if a set is convergent, divergent, or definitely divergent. I've seen the test to do in a textbook but i don't understand it.

    I'm supposed to find out what these 5 sets are, and find the limit if its convergent.
    I'd appreciate it if you showed me 1 or 2 of them as an example.
    Thank you

    a) an = n^2(n-3)

    b) 2n^3 + 4n^2 - 5 / 8n^3 - 2n

    c) n^2 / n^3 - 4

    d) (-n)^-3

    e) n(-1)^n
    Hi! I am not 100 % sure what you mean in every example. E.g. in a) do you mean n^{2(n - 3)} or n^2 (n - 3)? Anyway, both of these are definitely divergent for n \rightarrow \infty, since their absolute values are both bigger than n^2, which is unbound and divergent.

    And in b) do you mean:
    2 n^3 + 4 n^2 - \frac{5}{8 n^3} - 2 n,
    \frac{2 n^3 + 4 n^2 - 5}{8 n^3 - 2 n},
    2 n^3 + 4 n^2 - \frac{5}{8}n^3 - 2 n
    or something else? The first and the last definition is definitely divergent, because they are both bigger than 2 n^3, which is definitely divergent. The second definition converges, because it can be rewritten as:
    \frac{2 + 4/n - 5/n^3}{8 - 2/n^2}
    For n going to infinity all terms vanish except \frac{2}{8} = \frac{1}{4}.

    c) is also convergent regardless if you mean \frac{n^2}{n^3} - 4 = \frac{1}{n} - 4 or \frac{n^2}{n^3 - 4} = \frac{1/n}{1 - 4/n^3} \rightarrow \frac{0}{1} = 0.

    d) is convergent because for positive n: |(-n)^{-3}| = \frac{1}{n^3} \rightarrow 0

    e) is definitely divergent because for positive n: |n(-1)^n| = n \rightarrow \infty.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: March 29th 2010, 11:36 AM
  2. Replies: 2
    Last Post: August 4th 2009, 01:05 PM
  3. Replies: 3
    Last Post: April 6th 2009, 10:03 PM
  4. Replies: 8
    Last Post: February 21st 2009, 09:16 AM
  5. convergent or divergent?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 9th 2008, 08:14 AM

Search Tags


/mathhelpforum @mathhelpforum