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 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 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 .