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Innequality prove question

  1. Dec 21, 2008 #1
    Last edited by a moderator: Apr 24, 2017 at 9:37 AM
  2. jcsd
  3. Dec 21, 2008 #2


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    You need a little bit more. It is true that
    [tex]x- \frac{x^3}{3!}+ \frac{x^5}{5!}> x- \frac{x^3}{3!}[/tex]
    but you have to fit
    [tex]sin(x)= \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}[/tex]
    into that. Notice that because this is an alternating series with decreasing terms, the entire "tail" (all terms past "n") is less than the difference between the n-1 and n terms.
  4. Dec 21, 2008 #3
    what is the steps for the solution?
  5. Dec 21, 2008 #4
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