Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

CHAPTER 11 REVIEW 0 97
J'(xn)(xn - r) - f (xn) = R 1 (r). Dividing by f'(xn), we get Xn - r - ;,~:~) = :(~::).By the formula for Newton's
method, the left side of the preceding equation is Xn+I - r, so lxn+l - r l = I ;(~:}) I· Taylor's Inequality gives us
IR 1 (r)l ::=; I J'~~r)llr - xnl 2 • Comb i nin~ this inequality with the facts lf"(x)l ::=; M and IJ'(x)l ~/(gives us
lxn+l - rl ::=;
M \2
/( \xn - r ·
2
11 Review
CONCEPT CHECK
1. (a) See Definition 11.1.1.
(b) See Definition 11.2.2.
(c) The terms ofthe sequence {an} approach 3 as n becomes large.
(d) By adding sufficiently many terms of the series, we can make the partia l sums as close to 3 as we like.
2. (a) Sec the definition on page 721 [ET page 697].
(b) A sequence is monotonic if it is either increasing or decreasing.
(c) By Theorem 11.1.1 2, every bounded, monotonic sequence is convergent.
3. (a) See (4) in Section 11.2.
(b) The p-se~ies f: J.. is converge~ ! ifp > 1.
n=l 77,1'
4. If 2: a ,. = 3, then lim an = 0 and lim S n = 3.
n - oo
n ---t ex>
5. (a) Test for Divergence: If lim a,, does not exist or if lim an =I 0, then the series 2:"':.. 1
an is divergent.
n---+oo n --+ oo n-
(b) Imegral Test: Suppose f is a continuous, positive, decreasing function on [1, oo) and let an= f(n). Then the series
:2:;::'= 1
an is convergent if and only if the improper integral f 1
00
(i) If f 1
00
f(x) dx is convergent, then 2:;::'= 1
a, is convergent.
(ii) If f 1
00
f(x) dx is divergent, then I::;"= 1 a,. is divergent.
(c) Comparison Test: Suppose that E an and 2: b,. are series with positive terms.
(i) If E b,. is convergent and an ::=; bn for all n, then E a,. is also convergent.
(ii) If E br. is divergent and an ~ bn for a ll n, then 2: an is also divergent.
f(x) dx is convergent. In other words:
(d) Limit Comparison Test: Suppose that E a,. and E bn are series with positive tenns. If lim (a,/bn) = c, where c is a
n.- oo
finite number and c > 0, then either both series converge or both diverge.
(e) Alternating Series Test: I fthe alternating series :2:;::'= 1 ( - 1 )n- 1 bn = b1 - b2 + b:3 - b4 + bo - ba + · · · [b,. > OJ
satisfies (i) bn+l :::; bn for all n and (ii) lim b,. = 0, then the series is convergent.
n-