## 3.1. Sequences

### Proposition 3.1.6: Algebra on Convergent Sequences

Suppose and
are converging
to

*a*and*b*, respectively. Then- Their sum is convergent to
*a + b*, and the sequences can be added term by term. - Their product is convergent to
*a * b*, and the sequences can be multiplied term by term. - 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). - If
*a*for all_{n}b_{n}*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
*a _{n} a*, which
implies that there exists an integer

*N*such that

_{1}if| a_{n}- a | < / 2

*n > N*. Similarly, since

_{1}*b*there exists another integer

_{n}b*N*such that

_{2}if| b_{n}- b | < / 2

*n > N*. But then we know that

_{2}if| (a_{n}+ b_{n}) - (a + b) | = | (a_{n}- a) + (b_{n}- b) |

| a_{n}- a | + | b_{n}- b |

< /2 + /2 =

*n > max(N*, which proves the first statement.

_{1}, N_{2})
Proving the second statement is similar, with some added tricks.
We know that *{ b _{n} }* converges, therefore
there exists an integer

*N*such that

_{1}if| b_{n}| < |b| + 1

*n > N*. We also know that we can find integers

_{1}*N*and

_{2}*N*so that

_{3}if| a_{n}- a | < / (|b| + 1)

*n > N*, and

_{2}if| b_{n}- b | < / (|a| + 1)

*n > N*, because

_{3}*|a|*and

*|b|*are some

*fixed*numbers. But then we have:

if| a_{n}b_{n}- a b | = | a_{n}b_{n}- a b_{n}+ a b_{n}- a b |

= | b_{n}(a_{n}- a) + a (b_{n}- b) |

| b_{n}| |a_{n}- a | + | a | | b_{n}- b |

< (| b | + 1) / (|b| + 1) + | a | / (|a| +1) < 2

*n > max(N*, which proves the second statement.

_{1}, N_{2}, N_{3})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 thatWe now need to work out the contradiction: the idea is that sinceafor all_{n}b_{n}n, buta > b.

*a > b*there is some number

*c*such that

*b < c < a*.

Since<----------[b]-------[a]--------> <----------[b]--[c]--[a]-------->

*a*converges to

_{n}*a*, we can make the terms of the sequence fall between

*c*and

*a*, and the terms of

*b*between

_{n}*b*and

*c*. But then we no longer have that

*a*, which is our contradiction. Now let's formalize this idea:

_{n}b_{n}
Let *c = (a + b)/2*. Then clearly *b < c < a* (verify!).
Choose *N _{1}* such that

*b*if

_{n}< c*n > N*. That works because

_{1}*b < c*. Also choose

*N*such that

_{2}*a*if

_{n}> c*n > N*. But now we have that

_{2}forb_{n}< c < a_{n}

*n > max(N*. That is a contradiction to the original assumption that

_{1}, N_{2})*a*for all

_{n}b_{n}*n*. Hence it can not be true that

*a > b*, so that the statement is indeed proved.