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