Rate of Convergence

1. Jun 30, 2005

Zaphodx57x

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. Jun 30, 2005

Zaphodx57x

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. Jun 30, 2005

HallsofIvy

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. Jan 12, 2012

sgvaibhav

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