Solução_Calculo_Stewart_6e

F.
61. e s n
= e 1+ 1 2 + 1 3 +···+ 1 n = e 1 e 1/2 e 1/3 ···e 1/n > (1 + 1) 1+ 1 2
= 2 3 4
1 2 3
··· n +1
n
= n +1
1+
1
3 ···
1+
1
n
SECTION 11.2 SERIES ¤ 465
[e x > 1+x]
Thus, e s n
>n+1and lim
n→∞ es n
= ∞. Since{s n} is increasing, lim sn = ∞, implying that the harmonic series is
n→∞
divergent.
TX.10
63. Let d n be the diameter of C n . We draw lines from the centers of the C i to
the center of D (or C), and using the Pythagorean Theorem, we can write
1 2 +
1 − 1 2
2 d1 =
1+ 1 2
2 d1 ⇔
1=
1+ 1 2
2 d1 −
1 − 1 2
2 d1 =2d 1 [difference of squares] ⇒ d 1 = 1 . 2
Similarly,
1=
1+ 1 2
2 d2 −
1 − d 1 − 1 2
2 d2 =2d 2 +2d 1 − d 2 1 − d 1d 2
=(2− d 1)(d 1 + d 2)
⇔
d 2 = 1 (1 − d1)2
− d 1 = , 1=
1+ 1 2
2
2 − d 1 2 − d d3 −
1 − d 1 − d 2 − 1 2
2 d3 ⇔ d 3 =
1
d n+1 =
1 −
n
i=1 di 2
2 − n
i=1 d i
[1 − (d1 + d2)]2
, and in general,
2 − (d 1 + d 2 )
. If we actually calculate d 2 and d 3 from the formulas above, we find that they are 1 6 = 1
2 · 3 and
1
12 = 1
3 · 4 respectively, so we suspect that in general, d 1
n = . To prove this, we use induction: Assume that for all
n(n +1)
k ≤ n, d k =
1
k(k +1) = 1 k − 1
k +1 .Then n
i=1
2
1 − n
n +1
formula for d n+1 ,wegetd n+1 = =
2 −
n
n +1
d i =1− 1
n +1 =
1
(n +1) 2
n +2
n +1
=
n
n +1
[telescoping sum]. Substituting this into our
1
, and the induction is complete.
(n +1)(n +2)
Now, we observe that the partial sums n
i=1 d i of the diameters of the circles approach 1 as n →∞;thatis,
∞
a n = ∞
n=1
n=1
1
=1, which is what we wanted to prove.
n(n +1)
65. The series 1 − 1+1− 1+1− 1+··· diverges (geometric series with r = −1) so we cannot say that
0=1− 1+1− 1+1− 1+···.
67. ∞
n=1 ca
n = lim n
n→∞
i=1 ca i = lim c n
n→∞
i=1 a
i = c lim n
n→∞
i=1 a i = c ∞
n=1 a n, which exists by hypothesis.
69. Suppose on the contrary that (a n + b n ) converges. Then (a n + b n ) and a n are convergent series. So by Theorem 8,
[(an + b n ) − a n ] would also be convergent. But [(a n + b n ) − a n ]= b n , a contradiction, since b n is given to be
divergent.
71. The partial sums {s n } form an increasing sequence, since s n − s n−1 = a n > 0 for all n. Also, the sequence {s n } is bounded
since s n ≤ 1000 for all n. So by the Monotonic Sequence Theorem, the sequence of partial sums converges, that is, the series
an is convergent.