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. [tex] \lim_{n\rightarrow infinity} sin(1/n) = 0 [/tex] My thought would be to do the following [tex] |sin(1/n) - 0| <= 1 [/tex] But the book says to get a rate in the form [tex]1/n^p[/tex] The following also gives me trouble. [tex] \lim_{n\rightarrow infinity} sin(1/n^2) = 0 [/tex] which seems like it should converge faster than the the first one.
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 [tex]\lim_{n\rightarrow infinity} [ln(n+1) - ln(n)] = 0[/tex] I get to an answer of 2-n or so, maybe I should keep more terms. Anybody help would be appreciated.
Do you know that [tex]\lim_{x\rightarrow 0}\frac{sin(x)}{x} = 1[/tex]? If you let x= 1/n, that's the same as [tex]\lim_{n\rightarrow \infty}\frac{sin(1/n)}{1/n}= 1[/tex]. What does that tell you about the rate of convergence? To do sin(1/n^{2}), look at [tex]\frac{sin(1/n^2}{1/n^2}[/tex]