Proof: Let
. For all
,

and

Thus is decreasing and is increasing. Also, for all ,

so is bounded below by , and

so is bounded above by .

It follows that there exist real numbers and such that

Now

so .

It follows from the next lemma that ; i.e.,

Since for all

we have

and since

Thus, in all cases, ; i.e., approximates with an error of no more than .

Proof: Let be a precision function for and let be a precision
function for
. For all
, define

I claim is a precision function for , and hence . Let .

**Case 1:**- is even. Suppose is even.
Say where
. Then

**Case 2:**- is odd.
Suppose is odd. Say where
.
Then

and we will use these generalizations.

are decreasing positive null sequences, so

are summable; i.e.,

(These are the sequences we called and in example 10.3.)

Also,
,
with an error smaller than

. My calculator says

and

Hence, for all ,

i.e.,

Thus

Hence

for all .

If we can show that
is a null
sequence, it follows that

or in other words,

I claim is a null sequence for and hence (11.35) holds for . In particular,

First suppose , then for , so

Since is a null sequence for , it follows from the comparison test that is a null sequence for . Now suppose . Then

so and

If , then is a null sequence, so is a null sequence.

for all and using the ideas from the last example, show that

Conclude that

**a)****b)****c)**- (assume here ).