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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