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Rate of Convergence

  1. Jun 30, 2005 #1
    I'm not sure how to solve these problems. The example given in the book does not use trig functions. Any insight into how I solve these would be helpful.

    Find the following rates of convergence.
    \lim_{n\rightarrow infinity} sin(1/n) = 0
    My thought would be to do the following
    |sin(1/n) - 0| <= 1
    But the book says to get a rate in the form [tex]1/n^p[/tex]

    The following also gives me trouble.
    \lim_{n\rightarrow infinity} sin(1/n^2) = 0
    which seems like it should converge faster than the the first one.
  2. jcsd
  3. Jun 30, 2005 #2
    I made some progress by taking the maclaurin polynomial and only keeping the first couple terms. I can't get anything satisfactory for this one though
    [tex]\lim_{n\rightarrow infinity} [ln(n+1) - ln(n)] = 0[/tex]
    I get to an answer of 2-n or so, maybe I should keep more terms.
    Anybody help would be appreciated.
  4. Jun 30, 2005 #3


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    Do you know that [tex]\lim_{x\rightarrow 0}\frac{sin(x)}{x} = 1[/tex]? If you let x= 1/n, that's the same as [tex]\lim_{n\rightarrow \infty}\frac{sin(1/n)}{1/n}= 1[/tex]. What does that tell you about the rate of convergence?

    To do sin(1/n2), look at [tex]\frac{sin(1/n^2}{1/n^2}[/tex]
  5. Jan 12, 2012 #4
    im searching for tutorials on this section particularly....
    any links?
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