# Big O, little o demonstration

• March 16th 2011, 02:03 PM
arbolis
Big O, little o demonstration
I must demonstrate that if $\{ x_n \}$, $\{ y_n \}$ and $\{ \alpha _n \}$ are sequences of real number then the following is true:
if $x_n=o( \alpha _n)$ then $x_n=O(\alpha _n )$.

My attempt: I must show that knowing that $\lim _{n \to \infty} \frac{x_n}{\alpha _n}=0$ then there exist a constant $C$ and an integer $r$ such that $|x_n| for $n>r$.

But to me, any constant in R that I'd take would make $\lim _{n \to \infty} \frac{x_n}{\alpha _n} \neq 0$. For instance if the sequence $\alpha_n =x_n$, then $\lim _{n\to \infty} \frac{x_n}{\alpha _n}=\frac{1}{C} \neq 0$. In fact C would have to be infinite for the relation to be true.

Am I missing something or the problem asks me to show a wrong result?
• March 16th 2011, 02:12 PM
girdav
If you take $\alpha_n=x_n$, we don't have $x_n=o(\alpha_n)$ so we can't deduce the conclusion of the result is true.
The result is surely true because a convergent sequence is bounded.
• March 16th 2011, 06:50 PM
arbolis
You're right, I cannot choose my example.
To prove the afirmation, is my work valid?
What I must prove is equivalent to prove that $\lim _{n\to \infty} \left | \frac{x_n}{\alpha _n} \right | \leq$C. I take $C=0$ and I'm done?
• March 16th 2011, 07:33 PM
Jose27
Quote:

Originally Posted by arbolis
You're right, I cannot choose my example.
To prove the afirmation, is my work valid?
What I must prove is equivalent to prove that $\lim _{n\to \infty} \left | \frac{x_n}{\alpha _n} \right | \leq$C. I take $C=0$ and I'm done?

This looks odd, why don't you work with the basic definition of limits: $|x_n|<\varepsilon |\alpha _n|$ if $n>N$. Now pick a c such that $|x_m| if $0 and just take C the maximum of both.
• March 16th 2011, 07:58 PM
arbolis
Thanks Jose. Actually I think it would be overkill.
I want to correct my previous post into this one: The existance of a constant C and an integer r such that $|x_n| for n>r is equivalent to write $\lim _{n\to \infty} \left | \frac{x_n}{\alpha _n} \right |.
But I start with knowing that $\lim _{n\to \infty} \frac{x_n}{\alpha _n} =0$ and since $C \geq 0$ then the implication is obvious.

I said overkill in your base because from what I understand you want to prove it for all n while I think they ask for $n \to \infty$, unless I'm misunderstanding once again the exercise.
• March 16th 2011, 08:15 PM
Jose27
You're right, I missed the part where it said that only for sufficiently large n.
• March 16th 2011, 08:46 PM
arbolis
Quote:

Originally Posted by Jose27
You're right, I missed the part where it said that only for sufficiently large n.

They do not state "for sufficiently large n", but this is what I believe from the definitions I have for the little and big o notations. The problem is the one I wrote in the first post in the lines before "My attempt".

By the way I didn't know the result would hold for any n, I think this is a nice result and I thank you for pointing it out. :)