# Thread: Proving Convergence

1. ## Proving Convergence

Let (
an) be a sequence which converges to 0. Use the definition of a convergent sequence to prove that the sequence (an^2) to 0.

Definition: There exists an L that is Real s.t. for every E>0 there exists N that is a positive integer s.t. for every n that is a positive integer n>N implies that |an - L| < E

2. Originally Posted by ah-bee
Let (
an) be a sequence which converges to 0. Use the definition of a convergent sequence to prove that the sequence (an^2) to 0.

Definition: There exists an L that is Real s.t. for every E>0 there exists N that is a positive integer s.t. for every n that is a positive integer n>N implies that |an - L| < E

Since $\displaystyle a_n\to 0$ it is bounded, i.e. there is $\displaystyle A>0$ so that $\displaystyle |a_n|\leq A$.

This means, $\displaystyle |a_n^2| = |a_n||a_n|\leq A|a_n| < A\epsilon$ for $\displaystyle n>N$.

3. i dont understand how works. can u explain to me how it works?

4. Originally Posted by ah-bee
i dont understand how works. can u explain to me how it works?
There exists $\displaystyle A>0$ so that $\displaystyle |a_n|\leq A$. Because convergent sequences are bounded.

To prove that $\displaystyle a_n^2 \to 0$ we need to show $\displaystyle |a_n^2 - 0| < \epsilon$ for any $\displaystyle \epsilon > 0$ where $\displaystyle N$ is sufficiently large.

Give me any $\displaystyle \epsilon > 0$, then since $\displaystyle a_n \to 0$ it means $\displaystyle |a_n - 0| = |a_n|< \frac{\epsilon}{A}$ for $\displaystyle n>N$.

But then if $\displaystyle n>N$ we have $\displaystyle |a_n^2 - 0| = |a_n^2| = |a_n||a_n|\leq A|a_n| < A\cdot \frac{\epsilon}{A} = A$.

5. thanks, i understand it now, so if i were to get a question saying if an converges to k, prove an^2 converges to k^2, i would approach it in the same way?

6. Originally Posted by ah-bee
thanks, i understand it now, so if i were to get a question saying if an converges to k, prove an^2 converges to k^2, i would approach it in the same way?
Not exactly in the same way. But it should be similar.