Results 1 to 6 of 6

Math Help - Need help with convergence of this series

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    55

    Need help with convergence of this series

    Hi guys I've been doing some problems on series and I ran into this one,

    \sum _{n=1}^{\infty } \left( \int _{0}^{\frac{1}{n}}\!{\frac {<br />
\sqrt {x}}{1+{x}^{2}}}{dx} \right) <br />

    which I have no idea how to go about doing.

    Any help would be appreciated.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by Hweengee View Post
    Hi guys I've been doing some problems on series and I ran into this one,

    \sum _{n=1}^{\infty } \left( \int _{0}^{\frac{1}{n}}\!{\frac {<br />
\sqrt {x}}{1+{x}^{2}}}{dx} \right) <br />

    which I have no idea how to go about doing.

    Any help would be appreciated.

    Thanks.
    Think about what the integral looks like when n is large: then you integrate on a very small interval [0,\frac{1}{n}] the function \frac{\sqrt{x}}{1+x^2}, which is almost equal to \sqrt{x} since x is small... and you can integrate \int_0^{1/n}\sqrt{x}\,dx explicitly. It gives \frac{2/3}{n^{3/2}}, which is the general term of a convergent series.

    This justifies why the majoration \frac{\sqrt{x}}{1+x^2}\leq \sqrt{x} will allow you to conclude.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    Posts
    55
    So when n is large the interval [0, \frac{1}{n}] becomes small. And this implies that the area bounded by the curve becomes smaller and smaller, and so the series is likely to converge? Hence the comparison with a convergent series?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Hweengee View Post
    Hi guys I've been doing some problems on series and I ran into this one,

    \sum _{n=1}^{\infty } \left( \int _{0}^{\frac{1}{n}}\!{\frac {<br />
\sqrt {x}}{1+{x}^{2}}}{dx} \right) <br />

    which I have no idea how to go about doing.

    Any help would be appreciated.

    Thanks.
    You could also do it this way

    f(n)=\int_0^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^2}dx

    It should be clear that since the interval of integration is positive and so is the integrand that f(n)>0~\forall{n}\in\mathbb{N}

    We can also see that it is continuous (Ill try to prove this if you need be). And finally f(n)\in\downarrow. So we may apply the integral test

    Therefore the series shares the convergence/divergence of

    \int_1^{\infty}\int_0^{\frac{1}{n}}\frac{\sqrt{x}}  {1+x^2}dxdn

    Now switching the order of integration gives

    \int_0^1\int_1^{\frac{1}{x}}\frac{\sqrt{x}}{1+x^2}  dndx

    =\int_0^1\bigg[\frac{1}{\sqrt{x}(x^2+1)}+\frac{1}{\sqrt{x}}+\frac  {1}{2}\bigg]dx

    All of which converge (for the justification of the first just consider that \frac{1}{\sqrt{x}(x^2+1)}\sim\frac{1}{\sqrt{x}}. Therefore the series converges.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by Hweengee View Post
    So when n is large the interval [0, \frac{1}{n}] becomes small. And this implies that the area bounded by the curve becomes smaller and smaller, and so the series is likely to converge? Hence the comparison with a convergent series?
    Not only is the integral smaller and smaller, but it is like \frac{2/3}{n^{3/2}}. The idea is to neglect the denominator since it becomes close to 1, so that we obtain a function that we know how to integrate. Actually I almost wrote the full proof. Just use the majoration I gave and integrate it.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2008
    Posts
    55
    Ok thanks alot guys.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 3rd 2011, 01:12 AM
  2. Replies: 2
    Last Post: May 1st 2010, 09:22 PM
  3. Replies: 4
    Last Post: December 1st 2009, 03:23 PM
  4. Replies: 7
    Last Post: October 12th 2009, 10:10 AM
  5. series convergence and radius of convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 15th 2008, 08:07 AM

Search Tags


/mathhelpforum @mathhelpforum