Math 414
Professor Lieberman
March 3, 2003
HOMEWORK #6 SOLUTIONS
Section 2.5
√
8. (a) There are many different possibilities. The easiest one is an = n.
(b) Method 1: From Remark 2.1.8(e), we know that limn→∞ an+1 = limn→∞ an , so Theorem 2.2.1(a) implies that limn→∞ (an+1 − an ) = 0. Finally, Exercise 2.1.4 implies that
limn→∞ |an+1 − an | = 0.
Method 2: According to Theorem 2.5.9, (an ) is a Cauchy sequence, so, for every ε > 0,
there is a natural number n∗ such that, for all n, m ≥ n∗ , we have |an − am | < ε. Given
ε > 0, we note that n ≥ n∗ implies n + 1 ≥ n∗ , so, if n ≥ n∗ , we have |an − an+1 | < ε, and
therefore limn→∞ |an+1 − an | = 0.
Section 2.6
1. (c) b1 = 1/3, which is not an for any n, so (bn ) is not a subsequence of (an ).
2. (b) By direct evaluation, an = 0 if n is even, an = 1 if n = 4k + 1 for some integer k and
an = −1 if n = 4k − 1 for some integer k. It follows that the only subsequential limit points
are 0, −1, and 1. Therefore lim supn→∞ an = 1 and lim inf n→∞ an = −1.
(d) Now an = bn + cn , where bn = (−1)n and cn = −(−1)n /n. Therefore (an ) is the
sum of a divergent sequence and a convergent sequence, so it must diverge. It also follows
that the only subsequential limit points are −1 and 1 (because limn→∞ cn = 0 and the
only subsequential limit points for (bn ) are −1 and 1). Therefore lim supn→∞ an = 1 and
lim inf n→∞ an = −1.
7. Realize that you can’t use Exercise 2.5.4 directly because the sequence (an ) from that
exercise doesn’t have to be the same as the sequence (an ) in this exercise. To generate our
subsequence (bn ) = (af (n) ), we use induction. Take b1 = a1 . Then, once we have b1 , . . . , bk ,
we define bk+1 as follows. First, set εk = min{1/k, |s0 − a1 |, . . . , |s0 − af (k) }, where the dots
represent the differences |s0 − aj | with j taking on every integer value between 1 and f (k)
such that aj 6= s0 . Because s0 is an accumulation point of {an } (the set of values of the
sequence), there is a point x in the sequence such that x 6= s0 and |x − s0 | < εk . Therefore
x = am for some natural number m and we must have m > f (k) (because of our choice for
εk ). We then define f (k + 1) = m. This construction provides a strictly increasing function
f , so (bn ) is a subsequence of (an ). In addition, |bn − s0 | < 1/n, so (bn ) converges to s0 .
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