The Real And Complex Number Systems

∑
an x n converges uniformly on [0, 1] . Now, we give another proof of Abel’s
Limit Theorem as follows. Note that each term of ∑ a n x n is continuous
on [0, 1] and the convergence is uniformly on [0, 1] , so by Theorem 9.2, the
power series is continuous on [0, 1] . That is, we have proved Abel’s Limit
Theorem:
lim
x→1 − ∑
an x n = ∑ a n .
9.36 If each a n > 0 and ∑ a n diverges, show that ∑ a n x n → +∞ as
x → 1 − . (Assume ∑ a n x n converges for |x| < 1.)
Proof: Given M > 0, if we can find a y near 1 from the left such that
∑
an y n ≥ M, then for y ≤ x < 1, we have
M ≤ ∑ a n y n ≤ ∑ a n x n .
That is, lim x→1 −
∑
an x n = +∞.
Since ∑ a n diverges, there is a positive integer p such that
p∑
a k ≥ 2M > M. (*)
k=1
Define f n (x) = ∑ n
k=1 a kx k , then by continuity of each f n , given 0 < ε (< M) ,
there exists a δ n > 0 such that as x ∈ [δ n , 1), we have
n∑
a k − ε <
k=1
n∑
a k x k <
k=1
n∑
a k + ε (**)
k=1
By (*) and (**), we proved that as y = δ p
M ≤
p∑
a k − ε <
k=1
p∑
a k y k .
k=1
Hence, we have proved it.
9.37 If each a n > 0 and if lim x→1 −
∑
an x n exists and equals A, prove
that ∑ a n converges and has the sum A. (Compare with Theorem 9.33.)
Proof: By Exercise 9.36, we have proved the part, ∑ a n converges. In
order to show ∑ a n = A, we apply Abel’s Limit Theorem to complete it.
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