Proving a sequence is convergent.

[porroadventum]

1. The problem is if a_n is convergent then prove or disprove by giving a counter example that a_n² is also convergent.
2. Since a_n is convergent then for all ε>0 there exists n₀$\in$ $N$ such that la_n-Ll<ε for all n>=n₀

So I then tried squaring (a_n-L) which gives a_n² -2a_nL +L²<ε²
How do I manipulate this to show that a_n² has a limit L too?

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

 

[Mark44]

1. The problem is if a_n is convergent then prove or disprove by giving a counter example that a_n² is also convergent.



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

So I then tried squaring (a_n-L) which gives a_n² -2a_nL +L²<ε²
How do I manipulate this to show that a_n² has a limit L too?

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

If the sequence {a_n²} converges, then for any ε > 0, there is a number n₁ such that |a_n² - L₂ | < ε when n >= n₁. Given that the sequence {a_n} converges, can you use this to show that {a_n²} also converges?

 

[szynkasz]

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

 

[SammyS]

If the sequence {a_n²} converges, then for any ε > 0, there is a number n₁ such that |a_n² - L₂ | < ε when n >= n₁. Given that the sequence {a_n} converges, can you use this to show that {a_n²} also converges?

I'm quite sure Mark means, |a_n² - L² | < ε ...

 

[Mark44]

I'm quite sure Mark means, |a_n² - L² | < ε ...

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.

 

[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 $$n>max(N_1,N_2)$$