Results 1 to 4 of 4

Math Help - series convergence

  1. #1
    mms
    mms is offline
    Junior Member
    Joined
    Jul 2009
    Posts
    67

    series convergence

    study the convergence of the following series indicating the radius of convergence and the interval of convergence

    i) <br />
\sum\limits_{n = 1}^\infty {sen\left( {\pi \left( {n + \frac{1}{n}} \right)} \right)} x^n <br />

    ii) <br />
\sum\limits_{n = 1}^\infty {\ln \left( {1 + \frac{1}{{n^2 }}} \right)} x^n <br /> <br />

    please i need help with this to understand how to do it D:
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    16
    When you've written "sen", i believe you meant "sin". If that is so, you can try working on the function. You should check out the formula

    [LaTeX ERROR: Convert failed]

    After that, you gotta remember what the criteria for convergence of sums like [LaTeX ERROR: Convert failed] are. Also, what formulas involving [LaTeX ERROR: Convert failed] and limits do you have?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
     \displaystyle{\ln\left(1+\frac{1}{n^2}\right) = \ln\left(\frac{n^2+1}{n^2}\right) = \ln\left(n^2+1\right)-2\ln n}

    Let's use the ratio test:  \displaystyle{\lim_{n\to\infty}\left|\frac{a_{n+1}  }{a_n}\right| = |x|\cdot\lim_{n\to\infty}\left|\frac{\ln\left((n+1  )^2+1\right)-2\ln(n+1)}{\ln\left(n^2+1\right)-2\ln n}\right|}}

    (Applying L'Hopital's rule)  \displaystyle{= |x|\cdot\lim_{n\to\infty}\left|\frac{\frac{2(n+1)}  {(n+1)^2+1}-\frac{2}{n+1}}{\frac{2n}{n^2+1}-\frac2n}\right|} = |x|\cdot\lim_{n\to\infty}\left|\frac{\frac{2}{(n+1  )(n^2+2n+2)}}{\frac{2}{n(n^2+1)}}\right|}}

     \displaystyle{=|x|\cdot\lim_{n\to\infty}\left|\fra  c{n^3+n}{n^3+3n^2+4n+2}\right|}} = |x|\leq1

    That just leaves us to consider the end points:  x=\pm1 .

    When  x=-1 we apply the alternating series test to see it converges. I leave the case of  x=1 for you.

    So in the end we get the interval of convergence is  [-1,1] .
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Considering the identity...

    \displaystyle \sin \pi\ (n+\frac{1}{n}) = (-1)^{n} \sin \frac{\pi}{n} (1)

    ... and that for 'large' n is  \sin \frac{\pi}{n} \sim \frac{\pi}{n} , the two series...

    \displaystyle \sum_{n=1}^{\infty} \sin \{\pi\ (n+\frac{1}{n})\}\ x^{n}

    \displaystyle \sum_{n=1}^{\infty} (-1)^{n}\ \frac{x^{n}}{n} (2)

    ... have the same radious of convergence...

    For 'large n' is \ln (1+ \frac{1}{n^2}) \sim \frac{1}{n^{2}}, so that the two series...

    \displaystyle \sum_{n=1}^{\infty} \ln (1+\frac{1}{n^{2}})\ x^{n}

    \displaystyle \sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} (3)

    ... have the same radious of convergence...

    Kind regards

    \chi \sigma
    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: 3
    Last Post: December 1st 2009, 03:06 PM
  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