# Thread: Rates of convergence of sequences

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:

$\displaystyle \lim_{n \to \infty} sin{\frac{1}{n}} = 0$.

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

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

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

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

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 $\frac{1}{n}$ be inside the parenthesis, not just $n$?

2. To find the rate of convergence, you want to check this:

$\lim_{n \to \infty}{\left|\frac{\sin{(\frac{1}{n+1})}}{\sin{(\ frac{1}{n})}}\right|}$

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

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

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