Results 1 to 8 of 8

Math Help - r they convergent or not?

  1. #1
    Senior Member
    Joined
    Feb 2008
    Posts
    321

    r they convergent or not?

    Attached Thumbnails Attached Thumbnails r they convergent or not?-2.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi

    For the first one, you can try to show that \left|\frac{n^3+(-1)^n8n^2+1}{n^4+(-1)^{n+1}n^2}\right|\sim\frac{1}{n} for n\to\infty. For the second one, the ratio test may be helpful.

    Not completely off-topic question : If f \sim g do English speaking people say that g is an approximation of f ? Is there another way of saying this ?
    Last edited by flyingsquirrel; May 16th 2008 at 09:10 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Feb 2008
    Posts
    321

    sorry

    sorry, i found

    the first one is less or equal to (n^3+8*n^2+1)/(n^4-n^2),
    i think the later one is convergent. so the first one is convergent?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2008
    Posts
    77
    Quote Originally Posted by flyingsquirrel View Post
    Not completely off-topic question : If f \sim g do English speaking people say that g is an approximation of f ? Is there another way of saying this ?
    The only other way I can think of, off-hand, is that g approximates to f.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Quote Originally Posted by szpengchao View Post
    the first one is less or equal to (n^3+8*n^2+1)/(n^4-n^2),
    i think the later one is convergent.
    It's divergent. (and according to the hint I gave you, \sum \frac{n^3+(-1)^n8n^2+1}{n^4+(-1)^{n+1}n^2} can't be convergent) Anyway, the comparison test can also be used to show the divergence.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Feb 2008
    Posts
    321

    ok

    so how can u show that sequence tends to 1/n ????
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Quote Originally Posted by szpengchao View Post
    so how can u show that sequence tends to 1/n ????
    You can get it by showing that

    \frac{\left|\frac{n^3+(-1)^n8n^2+1}{n^4+(-1)^{n+1}n^2}\right|}{\frac{1}{n}}=\left|\frac{n^4+  (-1)^n8n^3+n}{n^4+(-1)^{n+1}n^2}\right|

    tends to 1 as n approaches \infty. (divide the numerator and the denominator by n^4)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by szpengchao View Post
    so how can u show that sequence tends to 1/n ????
    I actually just put a post up about this..

    If \lim_{x\to{c}}\frac{f(x)}{g(x)}=1

    Then you can say that as x\to{c} f(x)\sim{g(x)}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: May 2nd 2010, 04:25 AM
  2. Replies: 1
    Last Post: April 25th 2010, 09:46 PM
  3. Replies: 2
    Last Post: August 4th 2009, 02:05 PM
  4. Replies: 3
    Last Post: April 6th 2009, 11:03 PM
  5. Replies: 8
    Last Post: February 21st 2009, 10:16 AM

Search Tags


/mathhelpforum @mathhelpforum