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Thread: Rates of convergence of sequences

  1. #1

    Rates of convergence of sequences

    Apologies if this is in the wrong forum.

    I'm trying to find the rate of convergence of the following sequence:

    [TEX]\displaystyle \lim_{n \to \infty} sin{\frac{1}{n}} = 0[/TEX].

    To begin, I used knowledge of the Maclaurin series for sin (hopefully correctly):

    [TEX]\bigg|\sin{\frac{1}{n}} - 0 \bigg| = \bigg| \frac{1}{n} - \frac{\cos{\xi(\frac{1}{n})}}{6}\big(\frac{1}{n}\b ig)^3\bigg|[/TEX].

    However, there's already a big discrepancy between this attempt and the actual answer given, which starts:

    [TEX]\bigg|\sin{\frac{1}{n}} - 0 \bigg| = \bigg| \frac{1}{n} - \frac{\sin{\xi(n)}}{6}\big(\frac{1}{n}\big)^3\bigg |[/TEX].

    I'm confused for two reasons: from what I understand, the error term for the series should have a cos in it, not a sin. Furthermore, regardless of whether I'm using cos or sin, shouldn't [TEX]\frac{1}{n}[/TEX] be inside the parenthesis, not just [TEX]n[/TEX]?

  2. #2
    Junior Member
    Join Date
    Jan 2012
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    114
    To find the rate of convergence, you want to check this:

    [tex]\lim_{n \to \infty}{\left|\frac{\sin{(\frac{1}{n+1})}}{\sin{(\ frac{1}{n})}}\right|}[/tex]

    You may apply L'Hopital's Rule. First, use a change of variables (instead of the integer sequence, consider the functions, and let [tex]x[/tex] replace [tex]n[/tex]).

    [tex]\lim_{\stackrel{x \to n}{n \to \infty}}{\left|\frac{\sin{(\frac{1}{x+1})}}{\sin{( \frac{1}{x})}}\right|} \stackrel{\text{H}}{=} \lim_{\stackrel{x \to n}{n \to \infty}}{\left|\frac{-(x+1)^{-2}\cos{(\frac{1}{x+1})}}{-x^{-2}\cos{(\frac{1}{x})}}\right|}[/tex] [tex]= \lim_{\stackrel{x \to n}{n \to \infty}}{\left|\frac{x^2\cos{(\frac{1}{x+1})}}{(x^ 2+2x+1)\cos{(\frac{1}{x})}}\right|} = \lim_{n \to \infty}{\left|\frac{\cos{(\frac{1}{n+1})}}{(1+\fra c{2}{n}+\frac{1}{n^2})\cos{(\frac{1}{n})}}\right|} = 1[/tex]
    So, it converges sublinearly.

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