# Rate of Convergence

by Zaphodx57x
Tags: convergence, rate
 P: 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.
 P: 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.
 PF Patron Sci Advisor Thanks Emeritus P: 38,428 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 $$\frac{sin(1/n^2}{1/n^2}$$
P: 63

## Rate of Convergence

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