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

SECTION 11.2 SERIES 0 51
ak+l- a1.:+2 = ak+l- Hak+l + bk+l) = ~(ak+l
- bk+l) > 0, and
bk+l - bk+2 = bk+1 - Jak+lbk+l = ~ ( ~- Jak+l ) < 0 =* a 1.:+1 > ak+2 > bk+2 > bk+l•
so the assertion is true for n = k + 1. Thus, it is true for a ll n by mathematical induction.
(b) From part (a) we have a > a,. > an+ 1 > bn+ 1 > b.. > b, which shows that both sequences, {an} and { bn}, are
monotonic and bounded. So they are both convergent by the Monotonic Sequence Theorem.
(c) Let lim an =a an d I tm . b ,. =
{3
.
Th
en
li
m an+l = lim -a,.-+b.,
-- 2
n--+oo n--+oo n - oo n--+oo
=>
a: +f3
a:=--
2
=>
2a: = Q + {3 => Q = {3.
93. (a) Suppose {Pn} converges top. T hen Pn+l = ~
. a+p ...
=>
=>
bp
p=-­
a + p
=>
p 2 + ap = bp "* p(p + a - b) = 0 => p = 0 or p = b - a.
(b) Pn+l = ~ = (~);n < (!!.)Pn since 1 + Pn > 1.
a + pn 1 + ....2:: a a
a
(c) By part (b), P1 < (~)Po. P2 < ( ~ )P1 _< ( ~ y P Pn·
For n = 0, we have Pl - po = - -- - po =
bpo Po(b - a - Po)
> 0 since Po < b- a. So Pl > po.
a + Po a+po ·
Now we suppose the assertion is true for n = k, that is, Pk < b- a and Pk+l > Pk · Then
- b _ a_~ __ a(b - a)+ bp~.;- ap~.; - bpk __ a(b- a - Pk)
b- a - Pk+l __.!..____...!~ > 0 because p~; < b - a. So
a+pk a+pk a + pk
bpk+l Pk+l(b- a - Pk+l) .
Pk+l < b - a. And Pk+2 - Pk+l = - Pk+l = > 0 smce Pk+l < b - a. Therefore,
a + Pk+l
a+ Pk+l
Pk+2 > ·Pk+l· Thus, the assertion is true for n = k + 1. It is therefore true for all n by mathematical induction.
A similar proof by induction shows that ifpo > b- a, then Pn > b- a and {Pn} is decreasing.
In either case the sequence {Pn } is bounded and monotonic, so it is convergent by the Monotonic Sequence Theorem.
rt then follows from part (a) that lim Pn = b- a.
n-oo
11.2 Series
1. (a) A sequence is an ordered list of numbers whereas a series is the sum of a list of numbers.
(b) A series is convergent if the sequence of partial sums is a convergent sequence. A series is divergent if it is not convergent.
(i) 2012 Ccngagc L