Find the rates of convergence of the following sequences as n approaches infinit

Question

Find the rates of convergence of the following sequences as n approaches infinity:

lim(n to infinity) (sin 1/n)^2 = 0

lim(n to infinity) [ln(n+1)-ln(n)] = 0

The answers are (1/n^2) and (1/n) respectively, can you explain why this is to me?

Also:

Find the rates of convergence of the following functions as h approaches 0:

lim(h to 0) sinh/h = 1

lim(h to 0) (1-cosh)/h = 0

The answers are (h^2) and and (h), respectively, can you explain why this is to me as well?

Thank you so much! Please help! I have a big midterm in a few hours and I'm so confused on this section!!

Explanation / Answer

One way to determine the rate of convergence is to consult the associated Taylor series. (a) for (sin (1/n))^2, the Taylor series is (sin (1/n)) ^2 = (1/n - 1/3! 1/n^3 + ..... )^2 = 1/n^2 - 2/3! 1/n^4 + ....... As n -> infinity, the largest term is 1/n^2. (b) for ln(n+1) - ln(n) = ln( (n+1)/n ) = ln(1 + 1/n) the Taylor series is ln(1 + 1/n) = 1/n - 2/n^2 + 3/n^3 + .... and the largest term is 1/n. same game for sin(h)/h and (1- cos(h) )/h. Their Taylor series are sin(h)/h = (h - h^3/3! + ....)/h = 1 - h^2/3! so sin(h)/h -> 1 like h^2 (1-cos(h))/h = ( 1 - (1 - h^2/2! + .....) / h = h/2 so (1-cos(h))/h -> 0 like h.

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