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Convergement of Sin

  1. Mar 29, 2009 #1
    Determine whether this converges. If so, what number?

    http://texify.com/img/\LARGE\!\Sigma[/URL] [Broken] _{0}^{ \infty } \sin^n (\frac{ \pi }{4} %2B n \pi).gif[/PLAIN] [Broken]

    When I start plugging in values, I get :
    n= 0 f(n)=1
    n=1 f(n)= -\sqrt{2}/2
    n=2 f(n)= \sqrt{2}/2

    Using the formula, a/(1-r), I substitute and get 1/(1+1)=1/2. But when i look at the values in the table, it seems to approach 1.

    So does it approach 1 or 1/2 ?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 29, 2009 #2
    First, that formula for evaluating geometric series is only valid for -1 < r < 1, so your application to a ratio of -1 is incorrect.

    Second, your evaluation of the term itself for n=2 is incorrect. It should be 1/2, so the sum is 1 - 1/sqrt(2) + 1/2 thus far.

    Determine the form of the nth term first (with no sin involved). You should find that it leads to a familiar type of series.
     
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