## Proof of Theorem 1

PROOF OF THEOREM 1

It is easiest to prove (iii), then (ii), and finally (i).

(iii) The series and have radius of convergence r.

Let |x| < r. We may choose c with |x| < c < r. Then converges

absolutely. For positive infinite H, Theorem 1 in Section 9.1 (page 526) shows that |c/x|H/H is positive infinite, so H|x/c|H ≈ 0. Therefore

Then by the Limit Comparison Test, converges absolutely.

Similarly converges absolutely.

Now let |x| > r. Using the same test we can show that and

diverge. Therefore both series have radius of convergence r.

(ii)

Let 0 < c < r. Our proof has three main steps. First, get an error estimate for the difference between f(t) and the mth partial sum. Second, show that f(t) is continuous for -c ≤ t ≤ c. Third, show that f(t) has the required integral.

The series converges absolutely. Let Em be the tail

Then

limm→∞ Em = 0.

Moreover, for -c ≤ t ≤ c,

Therefore Em is an error estimate for f(t) minus the partial sum, (14)

We now prove f is continuous on [-c, c]. Since c was chosen arbitrarily between 0 and r, it will follow that f is continuous on (-r, r). Let t ≈ w in [-c, c]. For each finite m,

Therefore

st|f(t) - f(u)| ≤ Em + 0 + Em.

Since the Em's approach zero, it follows that f(t) ≈ f(u). Hence/is continuous on [-c, c].

To prove the integral formula we integrate both sides of Equation 14 from 0 to x. Let 0 < x.

Again since Em approaches zero, we conclude that

The case x < 0 is similar, (i)

Let

Integrating term by term,

Thus

By the Fundamental Theorem of Calculus, g(x) = f'(x).

In part (i) of the proof we needed part (iii) to be sure that the series for g(f) converges for - r < t < r, and part (ii) to justify the term by term integration.

Last Update: 2006-11-08