James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

appendix F Proofs of Theorems A47
Proof of 1 Suppose that o c n b n converges. Then, by Theorem 11.2.6, we have
lim n l ` c n b n − 0. According to Definition 11.1.2 with « − 1, there is a positive integer
N such that | c nb n | , 1 whenever n > N. Thus, for n > N, we have
| c nx n | − Z
c n b n x n
b n
n
Z − | c x
nb |Z n b
Z , Z x n
b
Z
If | x | , | b | , then | xyb | , 1, so o | xyb | n is a convergent geometric series. Therefore,
by the Comparison Test, the series | oǹ−N c nx n | is convergent. Thus the series o c nx n is
absolutely convergent and therefore convergent.
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Proof of 2 Suppose that o c n d n diverges. If x is any number such that | x | . | d | ,
then o c n x n cannot converge because, by part 1, the convergence of o c n x n would
imply the convergence of o c n d n . Therefore o c n x n diverges whenever | x | . | d | .
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Theorem For a power series o c n x n there are only three possibilities:
1. The series converges only when x − 0.
2. The series converges for all x.
3. There is a positive number R such that the series converges if | x | , R and
diverges if | x | . R.
ProoF Suppose that neither case 1 nor case 2 is true. Then there are nonzero numbers
b and d such that o c n x n converges for x − b and diverges for x − d. Therefore the set
S − hx | o c nx n convergesj is not empty. By the preceding theorem, the series diverges
if | x | . | d | , so | x | < | d | for all x [ S. This says that | d | is an upper bound for the
set S. Thus, by the Completeness Axiom (see Section 11.1), S has a least upper bound
R. If | x | . R, then x Ó S, so o c nx n diverges. If | x | , R, then | x | is not an upper
bound for S and so there exists b [ S such that b . | x | . Since b [ S, o c nx n converges,
so by the preceding theorem o c n x n converges.
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4 Theorem For a power series o c n sx 2 ad n there are only three possibilities:
1. The series converges only when x − a.
2. The series converges for all x.
3. There is a positive number R such that the series converges if | x 2 a | , R
and diverges if | x 2 a | . R.
ProoF If we make the change of variable u − x 2 a, then the power series becomes
o c n u n and we can apply the preceding theorem to this series. In case 3 we have con-
. R. Thus we have convergence for
vergence for | u | , R and divergence for | u |
| x 2 a | , R and divergence for | x 2 a | . R. ■
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