I'm doing some textbook problems in Burden and Faires Numerical Analysis, when I encountered this question that I did not understand how to do.
How would I go about finding the rate of convergence for
$$\lim_{n\to\infty}\sin\dfrac{3}{n}=0$$
I know that it is true, because $\dfrac{3}{n}$ goes toward $0,$ and $\sin0=0$, but I don't know the method/ procedure in order to calculate the order of convergence for this. Any help or tips would be appreciated. Thanks!
$\endgroup$ 12 Answers
$\begingroup$Let $x_n = \sin \frac{3}{n}$. We know $x_n \to 0$.
One approach is to calculate $\mu = \lim_n \frac{|x_{n+1}-L|}{|x_{n}-L|}$.
Here we have $\mu = \lim_n \left| \frac{ \sin\frac{3}{n+1}}{\sin \frac{3}{n}} \right| = \lim_n \left| \frac{ \frac{3}{n+1}}{ \frac{3}{n}} \right| = 1$, hence we have what is known as sublinear convergence.
$\endgroup$ $\begingroup$As you say, for small $x$ we have $\sin x \approx x$, so for large $n$ we have $\sin \frac 3n \approx \frac 3n$. Then whatever terminology you have for $\frac 3n$ approaching zero as $n$ goes to infinity.
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