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Proof of Theorem 1


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

(iii) The series 09_infinite_series-548.gif and 09_infinite_series-549.gif have radius of convergence r.

Let |x| < r. We may choose c with |x| < c < r. Then 09_infinite_series-550.gifconverges

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, 09_infinite_series-552.gif converges absolutely.

Similarly 09_infinite_series-553.gif converges absolutely.

Now let |x| > r. Using the same test we can show that 09_infinite_series-554.gif and

09_infinite_series-555.gif diverge. Therefore both series have radius of convergence r.


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 09_infinite_series-557.gif converges absolutely. Let Em be the tail



limm→∞ Em = 0.


Moreover, for -c ≤ t ≤ c,


Therefore Em is an error estimate for f(t) minus the partial sum, (14)09_infinite_series-560.gif

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,



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) 09_infinite_series-564.gif

Let 09_infinite_series-565.gif

Integrating term by term,



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