Rate of Convergence

  • Thread starter Zaphodx57x
  • Start date
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 [tex]1/n^p[/tex]

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.
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.


Science Advisor
Homework Helper
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/n2), look at [tex]\frac{sin(1/n^2}{1/n^2}[/tex]
im searching for tutorials on this section particularly....
any links?

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