1. ## rate of convergence

i've been stuck on this proof for hours:

suppose 0<p<q, a_n=a+O(n^(-p)) and b_n=b+O(n^(-q)) where O is the order of convergence as n approaches infinite. i need to prove that a_n*b_n=ab+O(n^(-p)), in another words that this function converges to ab with order of convergence n^(-p).

by definition, a sequence a_n converges to a with rate of convergence O(b_n) if there exists k such that abs (a_n-a)<=k*abs(b_n).
is it possible to prove (or even true) that abs (a_n*b_n-ab) <= abs (a_n-a)* abs (b_n-b)? this would allow us to find such a k.

thanks in advance for any help!

2. Originally Posted by squarerootof2
i've been stuck on this proof for hours:

suppose 0<p<q, a_n=a+O(n^(-p)) and b_n=b+O(n^(-q)) where O is the order of convergence as n approaches infinite. i need to prove that a_n*b_n=ab+O(n^(-p)), in another words that this function converges to ab with order of convergence n^(-p).

by definition, a sequence a_n converges to a with rate of convergence O(b_n) if there exists k such that abs (a_n-a)<=k*abs(b_n).
is it possible to prove (or even true) that abs (a_n*b_n-ab) <= abs (a_n-a)* abs (b_n-b)? this would allow us to find such a k.

thanks in advance for any help!
The point is that given the properties of big-O that:

$\displaystyle a_n b_n = ab +aO(n^{-q})+bO(n^{-p})+ O(n^{-q})O(n^{-p})$

and as $\displaystyle 0<p<q$ we are left with only the $\displaystyle O(n^{-p})$ term, so:

$\displaystyle a_n b_n = ab +aO(n^{-p})=ab+O(n^{-p})$.

Now if you are expected to prove this from the definition of big-O notation that is a different problem.

RonL

3. Originally Posted by CaptainBlack
The point is that given the properties of big-O that:

$\displaystyle a_n b_n = ab +aO(n^{-q})+bO(n^{-p})+ O(n^{-q})O(n^{-p})$

and as $\displaystyle 0<p<q$ we are left with only the $\displaystyle O(n^{-p})$ term, so:

$\displaystyle a_n b_n = ab +aO(n^{-q})=ab+O(n^{-p})$.

Now if you are expected to prove this from the definition of big-O notation that is a different problem.

RonL
i'm not convinced of your second statement. why would 0<p<q condition imply that we have a_n*b_n=ab+aO(n^(-q))=ab+O(n^(-p))?

4. Originally Posted by squarerootof2
i'm not convinced of your second statement. why would 0<p<q condition imply that we have a_n*b_n=ab+aO(n^(-q))=ab+O(n^(-p))?

There is a typo now corrected
RonL