Results 1 to 6 of 6

Math Help - Converging sequence.

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Converging sequence.

    \sum \frac{1}{n}sin \frac{n \pi}{2}

    Prove that this sequence converges.
    Tackling the sin \frac{n \pi}{2} first we get:

    when n=1 sin \frac{ \pi}{2}=1

    n=2 sin \pi=0

    n=3 sin \frac{3 \pi}{2}=-1

    n=4 sin 2 \pi=0

    When n is even the term being added onto the sequence is 0.

    sin \frac{n \pi}{2}=1 when n=1,5,.....,4n-3

    sin \frac{n \pi}{2}=-1 when n=3,7,..........,4n-1

    I have all this information but I don't know what to do with it!

    I was thinking of substituting 4n-3 and 4n-1 instead of sin \frac{n \pi}{2} but this only results in a divergent sequence. =(
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Well, that's a good point oO
    So note that it's 0 if n is even.
    Hence you can keep only the odd ones.

    So write n=2k+1

    Now, if k is even, it's in the form 4k'+1 and \sin \frac{n \pi}{2}=1=(-1)^{\color{red}even}
    and if k is odd, it's in the form 4k'-1 and \sin \frac{n \pi}{2}=-1=(-1)^{\color{red}odd}

    can you see where I want to come ?



    Try to see what \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} gives
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    I can and I have succesfully completed question 6 out of 17!!

    I would never have seen the substitution n=2k+1. It is a little ingenious.

    Quote Originally Posted by Moo View Post

    Try to see what \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} gives
    Right, we haven't been shown yet how to determine what it converges to. I can show you it converges though (which is a little easy, I know):


    \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}= -\sum_{k=0}^\infty \frac{(-1)^{k+1}}{2k+1}

    and by the alternating series test this converges since \frac{1}{2k+1} \rightarrow 0 \ as \ k \rightarrow 0

    So how do you figure out what it converges to?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Showcase_22 View Post
    I can and I have succesfully completed question 6 out of 17!!

    I would never have seen the substitution n=2k+1. It is a little ingenious.



    Right, we haven't been shown yet how to determine what it converges to. I can show you it converges though (which is a little easy, I know):


    \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}= -\sum_{k=0}^\infty \frac{(-1)^{k+1}}{2k+1}

    and by the alternating series test this converges since \frac{1}{2k+1} \rightarrow 0 \ as \ k \rightarrow 0
    and that it is decreasing ^^

    So how do you figure out what it converges to?
    I don't know, you're only asked to prove it converges

    I said "try to see what..." to give you which series you were looking for
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,657
    Thanks
    1615
    Awards
    1
    Look at the series for the \arctan(x) when x=1.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Showcase_22 View Post
    I can and I have succesfully completed question 6 out of 17!!

    I would never have seen the substitution n=2k+1. It is a little ingenious.



    Right, we haven't been shown yet how to determine what it converges to. I can show you it converges though (which is a little easy, I know):


    \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}= -\sum_{k=0}^\infty \frac{(-1)^{k+1}}{2k+1}

    and by the alternating series test this converges since \frac{1}{2k+1} \rightarrow 0 \ as \ k \rightarrow 0
    so what does the series converge to?
    Note that \forall{x}\backepsilon|x|<1~\frac{1}{1+x^2}=\sum_{  n=0}^{\infty}(-1)^nx^{2n}

    So notice that your sum is equivalent to

    \begin{aligned}\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}&=\sum_{n=0}^{\infty}\int_0^1(-1)^nx^{2n}dx\\<br />
&=\int_0^1\sum_{n=0}^{\infty}(-1)^nx^{2n}dx\\<br />
&=\int_0^1\frac{dx}{1+x^2}<br />
\end{aligned}

    EDIT: Plato beat me...sorry
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Converging/Diverging Sequence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 17th 2010, 02:29 PM
  2. Sequence converging to Zero.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 28th 2009, 09:41 AM
  3. converging sequence?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 29th 2009, 02:46 PM
  4. Converging Sequence
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 16th 2008, 10:55 AM
  5. Converging measurable sequence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 28th 2008, 12:56 PM

Search Tags


/mathhelpforum @mathhelpforum