# Proving a sequence is convergent.

1. Oct 5, 2012

1. The problem is if an is convergent then prove or disprove by giving a counter example that an2 is also convergent.

2. Since an is convergent then for all ε>0 there exists n0$\in$ $N$ such that lan-Ll<ε for all n>=n0

So I then tried squaring (an-L) which gives an2 -2anL +L22
How do I manipulate this to show that an2 has a limit L too?

Or should I be looking for a counter example? I can't think of any!

2. Oct 5, 2012

### Staff: Mentor

If the sequence {an2} converges, then for any ε > 0, there is a number n1 such that |an2 - L2 | < ε when n >= n1. Given that the sequence {an} converges, can you use this to show that {an2} also converges?

3. Oct 5, 2012

### szynkasz

You can also start from:
$$|a^2_n-L^2|=|a_n+L|\cdot|a_n-L|$$

4. Oct 5, 2012

### SammyS

Staff Emeritus
I'm quite sure Mark means, |an2 - L2 | < ε ...

5. Oct 5, 2012

### Staff: Mentor

No, I meant L2, to distinguish it from L. It might turn out that L2 = L2, but I didn't want to make that assumption.

6. Oct 7, 2012

### dirk_mec1

$$\forall \epsilon>0\ \exists N_1: n >N_1\ |a_n -L|< \epsilon$$

but also

$$\forall \epsilon>0\ \exists N_2: n >N_2\ |a_n -L|< \epsilon_1 = \frac{\epsilon}{M+L}$$

Remember that every convergent sequence is bounded from above, say in this case by M.

Now we get:

$$|a_n^2 -L^2| = |a_n-L||a_n+L| <\epsilon_1 (M+L)<\epsilon$$ if $$n>max(N_1,N_2)$$