Results 1 to 6 of 6

Math Help - convergence or divergence

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    39

    convergence or divergence

    sqrt(n^2+n)-n


    |cos(n*pi/2)+sin(n*pi/2|(2n-1)/(n+2)


    i can kind of start the first one and can't start the second one. any ideas?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    So do you not want help with the first one? I cannot understand what you mean at all in number 2. Please rewrite with more parentheses. I don't know what you mean by | |. Are those absolute value brackets or are they another form of grouping? Make a clear denominator and numerator, please. If there is more than one term being multiplied together, then use spacing between the terms to show that.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by needhelp101 View Post
    \sqrt{n^2+n}-n


    \frac{|\cos(\frac{n\pi}{2})+\sin(\frac{n\pi}{2})|(  2n-1)}{n+2}


    i can kind of start the first one and can't start the second one. any ideas?
    Is that what you mean? And are you referring to the sequences? In other words, \lim_{n\to\infty}\sqrt{n^2+n}-n and \lim_{n\to\infty}\frac{|\cos(\frac{n\pi}{2})+\sin(  \frac{n\pi}{2})|(2n-1)}{n+2}

    The first one's a classic example of a convergent limit involving \infty-\infty. Try multiplying by \frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n} which gives you \frac{n^2+n-n^2}{\sqrt{n^2+n}+n} = \frac{n}{\sqrt{n^2+n}+n}. Can you take it from here?

    Spoiler:

    \frac{n}{\sqrt{n^2+n}+n} = \frac{n}{n\sqrt{1+\frac{1}{n}}+n} = \frac{n}{n\left(\sqrt{1+\frac{1}{n}}+1\right)} = \frac{1}{\sqrt{1+\frac{1}{n}}+1}

    Letting n\to\infty, we have that this limit equals \frac{1}{\sqrt{1+0}+1} = \frac{1}{2}

    For the second one, break the limit up into \left(\lim_{n\to\infty}\left|\cos\frac{n\pi}{2}+\s  in\frac{n\pi}{2}\right|\right)\left(\lim_{n\to\inf  ty}\frac{2n-1}{n+2}\right).

    Calculate the limits separately. If one of them diverges, the whole thing does (unless one of them goes to zero, then you need to be careful). If both of them converge, the whole limit converges to the product of the two limits.

    Spoiler:

    Since \left|\cos\frac{n\pi}{2}+\sin\frac{n\pi}{2}\right| constantly fluctuates between 1 and -1, (it goes 1, 1, -1, -1, 1, 1, -1, -1...) this sequence diverges, as it will bounce back and forth between 2 and -2 (as the other limit converges to 2).
    Last edited by redsoxfan325; April 22nd 2009 at 07:27 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2009
    Posts
    39
    sorry about the confusion on the way i wrote the problem. yea, you got it right redsoxfa325 on how the problem should be written.

    for the first problem, i'm not really able to see what the next step would be.

    and for the second problem, i got that it diverges. is that right??
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2009
    Posts
    39
    the second problem is actually |cos(n*pi/2) + sin(n*pi/2)| * 2n-1/n+2.
    sorry about all this confusion.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    I added spoilers to my original post that contain solutions.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Convergence/divergence
    Posted in the Calculus Forum
    Replies: 9
    Last Post: March 30th 2011, 07:37 AM
  2. Convergence and Divergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 7th 2010, 12:57 AM
  3. Convergence and divergence
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 11th 2009, 07:36 PM
  4. convergence or divergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 23rd 2009, 05:42 PM
  5. please help with convergence divergence
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 2nd 2009, 05:46 AM

Search Tags


/mathhelpforum @mathhelpforum