Solve Rate of Convergence Problems Easily

[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.
<br /> \lim_{n\rightarrow infinity} sin(1/n) = 0<br />
My thought would be to do the following
<br /> |sin(1/n) - 0| &lt;= 1<br />
But the book says to get a rate in the form 1/n^p

The following also gives me trouble.
<br /> \lim_{n\rightarrow infinity} sin(1/n^2) = 0<br />
which seems like it should converge faster than the the first one.

 

[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.

 

[HallsofIvy]

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}

 

[sgvaibhav]

im searching for tutorials on this section particularly...
any links?