# Products of sequences

## 1. Properties of the notion of limit with multiplication

We saw in a previous page that limits work nicely with respect to the sum of two sequences. We prove an analogous result for multiplication here.

Theorem.

Let ( ) and ( ) be convergent sequences with limits and respectively. Let the sequence ( ) be defined by = . Then as .

Proof.

We are given that 0 - and 0 - and must prove that .

Let 0 be a bound for the sequence ( ) so for all ; this exists by the theorem on boundedness of convergent sequences. Also let =(, ) .

Subproof.

Let 0 be arbitrary.

Subproof.

Let 1 such that 1 - /2 and 2 such that 2 - /2 .

Let =( 1, 2) .

Subproof.

Let be arbitrary.

Subproof.

Assume .

Then 1 and 2 so - /2 and - /2 .

So ( )-( ) - + - = - + - /2+ /2= by the triangle inequality.

It follows that - .

So - .

So - .

So 0 - , as required.

## 2. The product of a bounded sequence and a null sequence

The next result is rather useful: it is a special case of the last result when one of the two sequences tends to zero. In this very special case, it doen't matter if the other sequence doesn't converge: it suffices that this other sequence is bounded. The proof is very similar.

Theorem.

Let ( ) and ( ) be two sequences and suppose that 0 as and that ( ) is a bounded (but not necessarily convergent) sequence. Let the sequence ( ) be defined by = . Then 0 as .

Proof.

Let 0 be a bound for the sequence ( ) so for all .

Subproof.

Let 0 be arbitrary.

Subproof.

Let such that / from the convergence of ( ) to 0.

Subproof.

Let be arbitrary.

Subproof.

Assume .

Then / and .

So = (/) =.

It follows that .

So .

So .

So 0 , as required.

You should note particularly that this last result say something special about the limit zero. There is no general result sating that the limit of the product of a convergent sequence ( ) and a bounded sequence ( ) exists. For a simple counterexample, take = to be the constant sequence and for the bounded divergent sequence (-1) . Then ( ) diverges for 0 .