1549901369-Elements_of_Real_Analysis__Denlinger_

8.6 Power Series 505

it is always an interval centered at c. The next theorem is the first step in

understanding this situation.

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Theorem 8.6.2 If a power series L ak(x - c)k converges for some x 1 =/:-c,

k=O

then it converges absolutely whenever Ix - cl < lx 1 - cl; that is, f or all x in the

interval (c - c, c + c), where c = lx 1 - cl.

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Proof. Suppose L ak(x1 - c)k converges for some X1 =/:-c. Let E: = lx 1 - cl

k=O

and choose any x E (c - E:, c + c). By the general term test, ak(x 1 - c)k---+ 0.

c-e

e

~

I I

c c +e

Figure 8.2

Hence { ak(x 1 - c)k} is a bounded sequence, so :JM> 0 such that \:/k,

lak(x1 - c)kl < M

lakl lx1 - elk< M.

Now, x E (c - E:, c + c), so Ix - cl < E:.

Thus \:/k, lak(x - c)kl = lakl Ix - elk= lakl lx1 - elk I x - c lk

<Mi x - e lk =Mrk,whererx~-1:-cl·

X1 - C X1 - C

Now 0 < r < 1 since r = t-ell < 1. Thus, I:, Mrk is a convergent

X1 - C

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(geometric) series. Therefore, by the comparison test,L lak(x-c)kl converges.

k=O

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Corollary 8.6.3 If a power series L ak(x - c)k does not converge absolutely

k=O

for some x 1 =/:-c, then it diverges whenever Ix - cl > lx1 - cl; that is, for all x

outside the interval [c - E:, c + c], where E: = lx1 - cl.

Proof. Exercise 1. •