12.8 Proof of the Differentiation Theorem

Proof: I'll show that for all .

- a) If converges, then converges.
- b) If converges and , then converges.

To prove b), suppose converges and . By lemma 12.73, converges, and hence is bounded. Choose such that

Then , and

By the comparison test, converges.

Proof: Use the lemma twice.

Proof: Let be a point in the disc of convergence, and let be a different point in the disc. Then

Let

Then

and since

the theorem will follow if we can show that is continuous at .

In the calculation below, I quietly use the following facts:

- a) When , .
- b) When , .

Let the radius of convergence of our power series be , and let . Then

Let . Then , and

(Here I've used the fact that for .) Thus

We noticed in the corollary to lemma 12.75 that the series has radius of convergence , and hence converges to a limit , and by (12.78),

If is a sequence in such that , then

for all large , and by the null-times bounded theorem and comparison theorem for null sequences, . Hence, is continuous at .