Question 9.T.16: Let R be the radius of convergence of the series ∑anx^n. The...
Let R be the radius of convergence of the series \sum{a_{n}x^{n}}. Then the sum S(x) = \Sigma^{∞}_{n=0} a_{n}x^{n} is differentiable on (−R, R), and its derivative is given by the series
S^{\prime}(x) = \sum\limits_{n=1}^{∞}{na_{n}x^{n−1}}, (9.20)
whose radius of convergence is also R.
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Since \lim n^{1/n} = 1, we can use Lemma 9.2 to obtain
\lim \sup |na_{n}|^{1/n} = \lim \sup|a_{n}|^{1/n} = 1/R.
By Theorem 9.15 both series \sum{a_{n}x^{n}} and \sum{na_{n}x^{n−1}} are uniformly convergent on [−r, r] for all 0 < r < R. Since any x ∈ (−R, R) lies in [−r, r] for some r ∈ (0, R), Equation (9.20) follows from Theorem 9.9.
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