# The Operator

For a differentiable function let be the function given by

Iterations of this operator'' yield many interesting identities.

## The Sum

For we have

On the other hand may be applied to the function to obtain

Comparing this with the above series expansion, we conclude that

Observe that after iterations, the resulting function takes the form

for some polynomial of degree It is therefore desirous to find the recurrence relation satisfied by the polynomials From the identity

(which, again, can be calculated by Maple,) we see that

holds for

# The Sum

For we have

On the other hand may be applied to the function to obtain

Comparing this with the above series expansion, we conclude that

• Use Maple to obtain the above expressions.

Observe that after iterations, the resulting function takes the form

for some polynomial of degree The recurrence relation satisfied by the polynomials may be deduced from the identity

we see that

# Absolute Convergent

The series is absolutely convergent if the series converges. (In more formal language, the sequence is absolutely summable if the sequence is summable.)

## Convergence of and

Theorem. Every absolutely convergent series is convergent. Moreover, a series is absolutely convergent if and only if the series formed from its positive terms and the series formed from its negative terms both converge.

Proof. If converges, then, by the Cauchy criterion,

Since

it follows that

which shows that converges.

To prove the second part of the theorem, let

so that is the series formed from the positive terms of , and is the series formed from the negative terms.

If and both converge, then

also converges, so converges absolutely.

On the other hand, if converges, then also converges. Therefore

and

both converge.

## Leibniz's Theorem.

Suppose that and that

and that

Then the series

converges.

Proof. We prove that

(1)

(2)

(3) if is even and is odd.

This is so since:

(1)

(2)

(3) If is even and is odd, choose such that

then

Now, the sequence converges, because it is nondecreasing and is bounded above (by for any odd ). Let

Similarly, let

If follows from (3) that since

it is actually the case that . This proves that

• Example: The series

is convergent, but not absolutely convergent.

# Uniform Convergence

DEFINITION: Let be a sequence of functions defined on , and let be a function which is also defined on Then is called the uniform limit of on if for every there is some such that for all in

We also say that converges uniformly to on , or that approaches uniformly on

THEOREM Suppose that is a sequence of functions which are integrable on and that converges uniformly on to a function which is integrable on . Then

Proof. Let There is some such that for all we have

Since this is true for any , it follows that

Theorem. Suppose that is a sequence of functions which are continuous on , and that converges uniformly on to . Then is also continuous on .

Proof. For each in we must prove that is continuous at We will deal only with in ; the cases and require the usual simple modifications.

Let . Since converges uniformly to on , there is some such that

In particular, for all such that is in , we have

(1)

(2)

Now is continuous, so there is some such that for we have

(3)

Thus, if , then

This proves that is continuous at

Suppose that is a sequence of functions which are differentiable on , with integrable derivatives , and that converges (pointwise) to Suppose, moreover, that converges uniformly on to some continuous function Then is differentiable and

Proof. Applying the previous theorem to the interval , we see that for each we have

Since is continuous, it follows that for all in the interval