Math 414
Professor Lieberman
February 24, 2003
EXAM #1 SOLUTIONS
1. the statement P (n) is “n5 − n is divisible by 5.” To check P (1), we compute 15 − 1 = 0,
which is divisible by 5 because 0 = 5 · 0. Now, we suppose that P (k) is true for some k ∈ N.
Then
(k + 1)5 − (k + 1) = k 5 − k + 5(k 4 + 2k 3 + k 2 )
by algebra, so (k + 1)5 − (k + 1) is also divisible by 5. Therefore P (k + 1) is also true.
Now that we have shown that P (1) is true and that P (k) implies P (k + 1), mathematical
induction implies that P (n) is true for any n ∈ N.
2. Given ε > 0, we want to find a natural number n∗ so that
√
n c − 1 < ε
√
√
if n ≥ n∗ . Since c > 1, we always have n c − 1 > −ε, so we only have to arrange n c − 1 < ε,
which is the same as (1 + ε)n > c. By the Archimedean axiom, there is an integer n∗ such
that n∗ ε > c − 1, and then n ≥ n∗ implies that 1 + nε > c. From Bernoulli’s inequality, we
see that (1 + ε)n ≥ 1 + nε > c.
3. A homework problem says that, if f and g are injections, then g ◦ f is also an injection,
and, if f and g are onto, then g ◦ f is onto. Therefore g ◦ f is a bijection. To see that
(g ◦ f )−1 = f −1 ◦ g −1 , we note that if c ∈ C with b = g −1 (c) and a = f −1 (b), then
f −1 ◦ g −1 (c) = f −1 (b) = a. But also g ◦ f (a) = c, so (g ◦ f )−1 (c) = a and therefore
(g ◦ f )−1 = f −1 ◦ g −1 .
4. There are several different ways to do this.
Method 1: Use the limit theorems:
2
3
2n + 3
2 + n3
lim 3
= lim n
= 0.
n→∞ n + 1
n→∞ 1 + 13
n
Method 2: Use the definition, along with knowing that the limit is zero. Given ε > 0,
take n∗ (by the Archimedean principle) so that n∗ > (3/ε)1/2 . If n ≥ n∗ , then
2n + 3
2n + 3
3n + 3
3
3
3
n3 + 1 − 0 = n3 + 1 ≤ n3 + 1 = n2 + n + 1 ≤ n2 ≤ (n∗ )2 < ε.
Method 3: Note first that (2n + 3)/(n3 + 1) ≥ 0 and that
2(n + 1) + 3
2n + 5
= 3
.
(n + 1)3 + 1
n + 3n2 + 3n + 2
By cross-multiplying, we have
(2n + 5)(n3 + 1) = 2n4 + 5n3 + 2n + 5 < 2n4 + 9n3 + 15n2 + 13n + 2 = (2n + 3)(n3 + 3n2 + 3n + 2),
so
2(n + 1) + 3
2n + 3
< 3
.
(n + 1)3 + 1
n +1
1
2
Hence we have a decreasing sequence, which is bounded below and therefore it converges.
Method 4: We have
2n + 3
3
0≤ 3
≤ 2.
n +1
n
(See the algebra in Method 2 to verify this last inequality.) But limn→∞ 0 = 0 and
limn→∞ 3/(n2 ) = 0, so
2n + 3
=0
lim
n→∞ n3 + 1
by the Squeeze Theorem.
√
5. Method 1: First, a1 = 2 ≤ 2 and, if an ≤ 2, then
√
√
an+1 = 2an ≤ 2 · 2 = 2,
so by induction an ≤ 2 for all n, so the sequence is bounded above. Also an ≥ 0 by the
formula, so
√
√
an+1 = 2an ≥ an an = an ,
so the
√ sequence is also increasing. It therefore
√ has a limit A. To evaluate A, we note that
A = √2A, so A = 0 or A = 2. Since a1 = 2, and (an ) is an increasing sequence, we have
A ≥ 2, so A = 2.
Method 2: The basic idea is to see that an = 2b(n) with b(n)
2−n . We proceed by
√= 1 −1/2
induction, or look at the pattern from the first few a’s: a1 = 2 = 2
and b(1) = 2−1 =
1/2. If we know that ak = 2b(k) for some k ∈ N, then
√
ak+1 = 2ak = 21/2 2b(k)/2 = 2(b(k)+1)/2 .
But (b(k) + 1)/2 = (1 − 2−k + 1)/2 = 1 − 2−(k+1) , so the formula for an is true by induction.
(Any verification of this formula by examining the first few a’s was worth few points.) Now
we note that b(n) → 1 as n → ∞, so
lim an = 2limn→∞ b(n) = 2.
n→∞
6. The set of accumulation points is [−2, 4]. It’s easy to show that points outside of this
interval are not accumulation points because if s0 is outside of this interval, then the
interval (s0 − ε, s0 + ε) has no points in S if ε = min{|s0 − (−2)|, |s0 − 4|}. For any point
s0 in the open interval (−2, 4), there is a number δ(s0 ) such that x ∈ S for all x with
0 < |x − s0 | < δ. (To be technically correct, you have to check the cases s0 = 1 and s0 6= 1
separately since δ = 3 if s0 = 1 and δ = min{|s0 − (−2)|, |s0 − 1|, |s0 − 4|} if s0 6= 1.) So the
number x = s0 +min{δ, ε/2} is in S with 0 < |x−s0 | < ε, which means s0 is an accumulation
point. For s0 = −2 and s0 = 4, a similar argument works with x = −2 + min{3, ε/2} if
s0 = −2 and x = 4 − min{3, ε/2} if s0 = 4. As mentioned in class, your argument did not
need anywhere near this much detail.