# Rate of Convergence

1. ### Zaphodx57x

31
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 $$1/n^p$$

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. ### Zaphodx57x

31
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
$$\lim_{n\rightarrow infinity} [ln(n+1) - ln(n)] = 0$$
I get to an answer of 2-n or so, maybe I should keep more terms.
Anybody help would be appreciated.

3. ### HallsofIvy

40,367
Staff Emeritus
Do you know that $$\lim_{x\rightarrow 0}\frac{sin(x)}{x} = 1$$? If you let x= 1/n, that's the same as $$\lim_{n\rightarrow \infty}\frac{sin(1/n)}{1/n}= 1$$. What does that tell you about the rate of convergence?

To do sin(1/n2), look at $$\frac{sin(1/n^2}{1/n^2}$$

4. ### sgvaibhav

68
im searching for tutorials on this section particularly....