for all , and claimed that a similar argument shows that

We will now use the basic properties of limits to prove (6.51). By the product rule and the null sequence rule,

Hence by the sum rule

By the sum rule and the null sequence rule

Hence by the product rule,

Now is a constant sequence, so by the product rule,

The argument given in equation (6.53) looks remarkably similar to the eighteenth century argument given in example 6.25.

In (2.13), we showed that

for all , and claimed that these inequalities show that . Now I want to examine the claim more closely.

In example 6.20 we proved that

and in example 6.50 we proved that

By applying the squeezing rule to equation 6.55, we see that

i.e.

Here I cannot apply the quotient rule for sequences, because the limits of the numerator and denominator do not exist. However, I notice that I can simplify my sequence:

I will now use a trick. I will factor the highest power of out of the numerator and denominator:

It is now clear what the limit is.

I'll apply the factoring trick of the previous example.

so

I observe that is a translate of so by the translation rule

I can also try to do this by my factoring trick:

- a)
- b)
- c)
- d)
- e)
- f)
- g)
- h)
- .

In (2.34) we showed that

I want to conclude from this that .

By the th root rule, and the quotient and product rules, we have

and

By (6.61) and the squeezing rule, we conclude that

i.e.

Thus, for example

and

It is clear that for all in . Let . Then by the translation rule, also. From (6.63) we have

Thus

i.e.

Hence

Thus , and it follows that or . But we noticed above that for all in , and hence by the inequality rule for sequences, . Hence we conclude that , i.e.,

(Actually there is an error in the reasoning here, which you should try to find, but the conclusion (6.64) is in fact correct. After you have done exercise 6.68, the error should become apparent.)

Thus, for example

and

Notice that for all . Let

By the translation rule

By (6.67) (with replaced by ), we have

Hence

Thus

By the quadratic formula

Since for all , we have , so we have

(This example has the same error as the previous one.)

- a)
- Let be a convergent sequence of real numbers. If for all in , then .
- b)
- Let and be real sequences. If , then .
- c)
- Let be a real sequence. If then either or .
- d)
- Let and be real sequences. If , then either or .

In (2.4) we showed that

Use this result, together with the power sums listed in Bernoulli' table to find the area of .

Proof: Let . Now is a translate of , so by the translation theorem

so we have or

We assumed that , so , and hence it follows that .

The proof just given is not valid. In fact, the argument shows that whenever , and this is certainly wrong when . The error comes in the first sentence, `` Let ''. The argument works if the sequence or converges. We will now give a second (correct) proof of theorem 6.71.

Second Proof: Let be a real number with .
If , then
is a constant sequence, and
. Hence the theorem holds when ,
and we may assume that .
Let be a generic positive number,
If
we have

Now since , we know that and hence

By the Archimedean property, there is some positive integer such that . Then for all in

Hence .