# 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 $l ∈ ℝ$ and $m ∈ ℝ$ 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 and 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 $n ∈ ℕ$; 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 $N 1 ∈ ℕ$ such that and $N 2 ∈ ℕ$ such that .

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

Subproof.

Let $n ∈ ℕ$ 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 .

So .

So , 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 $n ∈ ℕ$.

Subproof.

Let $ε > 0$ be arbitrary.

Subproof.

Let $N ∈ ℕ$ such that from the convergence of $( a n )$ to $0$.

Subproof.

Let $n ∈ ℕ$ 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 .

So .

So , 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 $l ≠ 0$.