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Convergence/divergence of series

  1. Nov 7, 2005 #1
    \sum\limits_{n = 1}^\infty {\frac{{\sin \left( {\frac{{n\pi }}{2}} \right)}}{{11 + 8n}}}

    I know that the numerator oscillates between -1 and 1 but there are some values of n for which the sine term also takes on the value of zero. So I can't find an explicit form for the numerator which means I can't use the alternating series test. I can't think of any other tests to use since the expression inside the summation takes on negative values 'regularly.'

    \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^n \frac{1}{{n^{1 + \frac{1}{n}} }}}

    Hmm...this one is a bit trick so basically I just hoped that the alternating series test would yield something simple.

    n^{1 + \frac{1}{n}} \ge n^{1 + \frac{1}{{n + 1}}} \Leftrightarrow \frac{1}{{n^{1 + \frac{1}{n}} }} \le \frac{1}{{n^{1 + \frac{1}{{n + 1}}} }}

    a_n \le a_{n + 1}

    The terms are not decreasing so the series diverges? My caculator suggests otherwise. Again, I'm not sure about this one.

    Can someone help me out with these two series?
  2. jcsd
  3. Nov 7, 2005 #2


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    You can just ignore the 0 terms. If you want be precise, since [itex]sin(\frac{n\pi}{2})[/itex] is 0 whenever n is even, replace n with 2m+1. The series is exactly the same as
    [tex]\sum\limits_{m = 0}^\infty {\frac{{\sin \left( {\frac{{(2m+1)\pi }}{2}} \right)}}{{11 + 8(2m+1)}}} [/tex]
    That "skips" the 0 terms and is an alternating series with decreasing terms (in absolute value) and so converges.
  4. Nov 7, 2005 #3
    For the second series, finding the limit of the An may help.

    This combined with the fact that the series is alternating may say something..
    Last edited: Nov 7, 2005
  5. Nov 7, 2005 #4
    Thanks for the help guys. I'll try to finish them off now.
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