3.1. Sequences

Proposition 3.1.6: Algebra on Convergent Sequences

Suppose and are converging to a and b, respectively. Then
  1. Their sum is convergent to a + b, and the sequences can be added term by term.
  2. Their product is convergent to a * b, and the sequences can be multiplied term by term.
  3. Their quotient is convergent to a / b, provide that b # 0, and the sequences can be divided term by term (if the denominators are not zero).
  4. If an bn for all n, then a b

Proof:

The proofs of these statements involve the triangle inequality, as well as an occasional trick of adding and subtracting zero, in a suitable form. A proof of the first statement, for example, goes as follows.

Take any > 0. We know that an a, which implies that there exists an integer N1 such that

| an - a | < / 2
if n > N1. Similarly, since bn b there exists another integer N2 such that
| bn - b | < / 2
if n > N2. But then we know that
| (an + bn) - (a + b) | = | (an - a) + (bn - b) |
      | an - a | + | bn - b |
      < /2 + /2 =
if n > max(N1, N2), which proves the first statement.

Proving the second statement is similar, with some added tricks. We know that { bn } converges, therefore there exists an integer N1 such that

| bn | < |b| + 1
if n > N1. We also know that we can find integers N2 and N3 so that
| an - a | < / (|b| + 1)
if n > N2, and
| bn - b | < / (|a| + 1)
if n > N3, because |a| and |b| are some fixed numbers. But then we have:
| an bn - a b | = | an bn - a bn + a bn - a b |
      = | bn(an - a) + a (bn - b) |
      | bn| |an - a | + | a | | bn - b |
      < (| b | + 1) / (|b| + 1) + | a | / (|a| +1) < 2
if n > max(N1, N2, N3), which proves the second statement.

The proof of the third statement is similar, so we will leave it as an exercise.

The last statement does require a new trick: we will use a proof by contradiction to get that result:

Assume that an bn for all n, but a > b.
We now need to work out the contradiction: the idea is that since a > b there is some number c such that b < c < a. 
 
<----------[b]-------[a]--------> 
 
<----------[b]--[c]--[a]-------->
Since an converges to a, we can make the terms of the sequence fall between c and a, and the terms of bn between b and c. But then we no longer have that an bn, which is our contradiction. Now let's formalize this idea:

Let c = (a + b)/2. Then clearly b < c < a (verify!). Choose N1 such that bn < c if n > N1. That works because b < c. Also choose N2 such that an > c if n > N2. But now we have that

bn < c < an
for n > max(N1, N2). That is a contradiction to the original assumption that an bn for all n. Hence it can not be true that a > b, so that the statement is indeed proved.

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