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

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

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 $A>0$ so that $|a_n|\leq A$. Because convergent sequences are bounded.

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

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

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

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.