508 Chapter 8 • Infinite Series of Real NumbersThe root test (8.2.15) may also be used in finding the interval of convergence
of a power series. In fact, the upper limit form of the root test applies more
generally because it does not require the existence of the limit.00
Theorem 8.6.8 The radius of convergence of the power series L ak(x - c)k
k=O
1 - k/f::":I
is p = R' where R = lim y lakl· (p = 0 if R = +oo, and p = +oo if R = 0.)
k->oo
Proof. Exercise 3. •The reader has surely had experience finding the interval of convergence of
power series in elementary calculus courses, so no additional examples will be
worked here. More examples can be found in Exercise Set 8.6.The next theorem will be no surprise. You probably already expect it.00
Theorem 8.6.9 (Algebra of Power Series) Suppose f(x) = I: ak(x-c)k
k=O
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and g(x) = L bk(x - c)k, with radii of convergence Pt and Pg respectively. Let
k=O
p = min{pt, pg}· Then, for all x in (c - p, c + p) and all t E IR,00
(a) tf(x) = L tak(x - c)k,
k=O
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(b) f(x) + g(x) = L (ak + bk)(x - c)k,
k=O
00 00
(c) f(x)g(x) = L ck(x-c)k, where ck= L ajbk-j as in the Cauchy product
00 k=O 00 k=O
of L ak and I: bk.
k=O k=OProof. Exercise 5. •POWER SERIES AS FUNCTIONS
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It is clear that a power series L ak(x-c)k may be regarded as a function of
k=O
x. From Corollary 8.6.4 we know that the domain of this function is a nonempty
interval centered at c, and may or may not include the endpoints of this interval
if it has any. Naturally, we are concerned whether this function is differentiable
on this interval. If so, it will also be continuous and integrable there.