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 ( a n ) and ( b n ) be convergent sequences with limits l and m respectively. Let the sequence ( c n ) be defined by c n = a n · b n . Then c n l · m as n .

Proof.

We are given that ε > 0   N   n   n N | a n - l | < ε and ε > 0   N   n   n N | b n - m | < ε and must prove that a n b n l m .

Let M > 0 be a bound for the sequence ( b n ) so | b n | < M for all n ; this M exists by the theorem on boundedness of convergent sequences. Also let K = max ( M , | l | ) .

Subproof.

Let ε > 0 be arbitrary.

Subproof.

Let N 1 such that n   n N 1 | a n - l | < ε / 2 K and N 2 such that n   n N 2 | b n - m | < ε / 2 K .

Let N = max ( N 1 , N 2 ) .

Subproof.

Let n be arbitrary.

Subproof.

Assume n N .

Then n N 1 and n N 2 so | a n - l | < ε / 2 K and | b n - m | < ε / 2 K .

So | ( a n b n ) - ( l m ) | | a n b n - l b n | + | l b n - l m | = | b n | | a n - l | + | l | | b n - m | < K ε / 2 K + K ε / 2 K = ε by the triangle inequality.

It follows that n N | a n b n - l m | < ε .

So n   n N | a n b n - l m | < ε .

So N   n   n N | a n b n - l m | < ε .

So ε > 0   N   n   n N | a n b n - l m | < ε , 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 ( a n ) and ( b n ) be two sequences and suppose that a n 0 as n and that ( b n ) is a bounded (but not necessarily convergent) sequence. Let the sequence ( c n ) be defined by c n = a n · b n . Then c n 0 as n .

Proof.

Let M > 0 be a bound for the sequence ( b n ) so | b n | M for all n .

Subproof.

Let ε > 0 be arbitrary.

Subproof.

Let N such that n   n N | a n | < ε / M from the convergence of ( a n ) to 0 .

Subproof.

Let n be arbitrary.

Subproof.

Assume n N .

Then | a n | < ε / M and | b n | < M .

So | a n b n | = | a n | | b n | < ( ε / M ) M = ε .

It follows that n N | a n b n | < ε .

So n   n N | a n b n | < ε .

So N   n   n N | a n b n | < ε .

So ε > 0   N   n   n N | a n b n | < ε , 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 ( a n ) and a bounded sequence ( b n ) exists. For a simple counterexample, take a n = l to be the constant sequence and for b n the bounded divergent sequence ( -1 ) n . Then ( a n b n ) diverges for l 0 .