MATH MATH 131A, FALL 2006 HOMEWOR HOMEWORK K 6 SOLUTIONS SOLUTIONS
3.3, 3.3, ex. 2: Clai Claim m 1: The sequence is bounded below by 1, i.e., i.e., x x n > 1, > 1, for all n all n.. This is true for
n = 1, by assumption. Assume xk > 1 for some k some k.. Then 1/x 1/xk < 1, so xk+1 = 2
− x1k > 2 − 1 = 1.
Claim 1 follows by the PMI. Claim 2: xn > xn+1 , for all n all n.. For n = 1, we have x2 = 2
− x1
< x1 ,
1
since this inequality is equivalent to the true inequality x inequality x 21 some k some k.. Then 1/x 1/xk+1 > 1 > 1/x /xk , so xk+2 = 2
− xk1
< 2
+1
− 2x
− x1k = xk
1
+1
+ 1 > 0. > 0. Assume x Assume x k+1 < xk , for
.
Claim 2 follows by the PMI. Since (x (xn ) is decreasing and bounded, it follows that it converges, say to x to x.. Letting n in xn+1 = 2 x1 yields
→∞
x = 2
−
− x1
n
x2
⇔
− 2x + 1 = 0.0.
Therefore, x = 1.
3.3, 3.3, ex. 4: Clai Claim m 1: (xn ) is increasing, i.e., xn
∈ N. √ For n = 1,√ this is true, , for some k some k.. Then xk = 2 + xk ≤ 2 + xk =
√ = 1 < 3 = x . Assume x Assume x ≤ x
≤ xn
+1
, for all n
since x since x 1 2 +1 +1 k k+1 xk+2 . By the Principle of Mathematical Induction (PMI), it follows that (x (xn ) is increasing. Claim 2: xn < 2, for all n Assume me xk < 2, for N. For n = 1, this is true since x1 = 1. Assu some k some k.. Then xk+1 = 2 + xk < 2 + 2 = 2. By the PMI, claim 2 is true. Since (x (xn ) is bounded above and increasing, it follows that it is convergent. Let x = limn xn . Then, by letting n letting n in x in x n+1 = 2 + xn (recall that x that x n+1 x, since (x (xn+1 ) is a subsequence of (xn )), we obtain
√
∈ √
√
→∞
→∞
→
x =
√ 2 + x.
Squaring both sides and moving all the terms to the left-hand side, we obtain x2
− x − 2 = 0. The solutions to this quadratic equation are −1 and 2. x ≥ 1, so x so x = = 2. Thus, x Thus, x n → 2, as n as n → ∞.
Sinc Sincee xn
≥ 1, for all n, it follows that
3.3, 3.3, ex. 9: We construct a sequence (x (xn ) with the desired properties inductively . Since u Since u = = sup A,
there exists a A such that u 21 < a u. (Obser (Observe ve that that a = u is allowed allowed.) .) Define x1 = a. Suppose that x that x 1 , . . . , xn have been constructed so that x that x 1 xn , xk A, and
∈
−
≤
u
− 21k < x k ≤ u, 1
≤ · · · ≤
∈
2
for all 1 k n. n . We now define define xn+1 as follows. follows. Let y = max u 2 1+1 , xn . If y = u, u , define xn+1 = y = y.. If y y < u, then there exists z exists z A such that y that y < z u. We set x set x n+1 = z = z.. Observe that 1 xn xn+1 , xn+1 A and u 2 +1 < xn+1 u. This defines the sequence (x (xn ) inductively inductively.. By construction, construction, (x (xn) is increasing, in A, and it converges to u to u..
≤ ≤
≤
∈
−
∈
≤
≤
n
{ −
}
n
3.3, 3.3, ex. 12: Since
xn+1
− xn = (n +1 1)
> 0, 0 ,
2
it follows that (x (xn ) is (strictly) (strictly) increasing. increasing. Furthermore urthermore,, n
xn = 1 +
k=2 n
< 1 +
k=2 n
= 1+
1 k2
− − − − − − 1
k (k
k=2
1)
1
1 k
1
k
1 = 1+ 1 2 1 =1+1 n < 2. 2 .
1 2
+
1 3
1 3
+
1 4
+
···+
1
n
1
− 2 − n − 1
+
1
n
− 1 −
1 n
−
Therefore, xn < 2, < 2, for all n all n
≥ 2. Since (x (xn ) is increasing and bounded above, it converges.
Remark: The limit of this sequence equals π2 /6, but this is not easy to prove. prove. The sum of the
above type, n
(ak
k=1
− ak
+1
) = a 1
− an ,
is called a telescopic sum. 3.3, 3.3, ex. 13: (a) We have: n+1
n
1 1+ n
1 1+ n
=
n+1 n
1
→e
= e.
(b) Similarly, 2n
n
2
→ 1 1+ n
=
1 1+ n
e2 .
(c) Analogously,
1 1+ n+1
n
=
1 1+ n+1
n+1
n n+1
1
→e
= e.
3
(d) We have: n
n
− − 1
1 n
n
=
1
n 1
=
n
n
n−1
1
=
1+
n
1
n−1
1
=
1+
→ 1e .
n−1
1
n n−1
n−1
3.4, 3.4, ex. 5: (
are X and Y and Y as subsequences of Z , ⇐) If Z is Z is convergent, then so are X Z , and they have the
same limit.
( ) Suppose xn a and yn a. Let ε > 0 be arbitrary. arbitrary. Then there exist exist K, L N such that n K xn a < ε and n L yn a < ε. Let M = = max 2K 1, 2L and assume n M . M . If n = 2k 1 is odd, then zn = x k and k K , so zn a < ε. If n = 2k is even, then z then z n = y 2k and k L, so zn a < ε. Therefore, zn a, as n .
⇒
| − | →∞
→ → ≥ ⇒ ⇒ | − | { − }
∈
≥ ⇒| − | − ≥ | − |
≥
n
≥ →
1 + n1 . Since an e, it follows that a2n e, as n subsequenc subsequencee of a conver convergent gent sequence sequence conver converges ges to the same limit). Therefore, Therefore,
→ → →
3.4, 3.4, ex. 7: (b) Let an =
2 1+ n
n
= a 1n/2
1 1+ n/2 n/2
=
→ ∞ (the
n
1 1+ 2n
(d) Similarly,
→
e1/2 .
n/2
2
e2 .
(an ) (which exists by assumption) by a. 3.4, 3.4, ex. 11: Let an = ( 1)n xn and denote the limit of (a
−
Since every subsequence of ((a an ) must converge to the same limit as the whole sequence, we obtain: a = lim a2k = lim x2k 0, k →∞
k→∞
≥
a = lim a2k+1 = lim ( x2k+1 ) k→∞
k→∞
−
≤ 0.
Therefore, a = 0. It follows that the sequence x sequence x n = an is also convergent and
| |
xn = an
| | → |a| = 0. 3.4, 3.4, ex. 12: Since (x (xn) is unbounded, no k ∈ N is an upper bound for the sequence. sequence. for each k there exists nk ∈ N such that x that x n > k. Therefore,
Therefore, Therefore,
k
0 <
1 1 < , xn k k
so by the Squeeze Theorem, 1/x 1/xn
k
→ 0, as k → ∞.
