Math 2260 Written HW #10 Solutions
1. Consider the sequence
1 1 3 5 7 9
− , , , , , ,... .
2 5 10 17 26 37
What is the nth term in this sequence?
Answer: Notice that the sequence of numerators is
−1, 1, 3, 5, 7, 9, . . .
which is just the odd numbers starting from −1. In fact, we can recognize that the nth term
in this sequence of numerators is 2n − 3.
On the other hand, the sequence of denominators is
2, 5, 10, 17, 26, 37, . . .
which doesn’t look like anything in particular, but if you subtract one from each term you
get
1, 4, 9, 16, 25, 36, . . .
which is obviously the sequence of perfect squares. Therefore, the nth term in the sequence
of denominators is n2 + 1.
Putting this together, then, the nth term of the given sequence is
2n − 3
.
n2 + 1
2. Consider the sequence (an )∞
n=1 where
an =
n!
.
nn
Does the sequence converge or diverge? If it converges, find the limit. If it diverges, explain
why.
Answer: First, let me write out an slightly more explicitly:
an =
n · (n − 1) · (n − 2) · · · · · 3 · 2 · 1
,
n · n · n · ··· · n · n · n
where there are exactly n factors in both the numerator and the denominator. Re-writing
slightly, I see that
an =
n · (n − 1) · (n − 1) · · · · · 3 · 2 1
1
1
· <1· = .
n · n · n · ··· · n · 2
n
n
n
On the other hand, an is always positive, so I have that
0 < an <
1
1
n
for all n.
Now, the sequence (0, 0, 0, 0, 0, 0, . . .) certainly converges to zero, and
1
=0
n→∞ n
lim
as well. Therefore, the sequence (an )∞
n=1 is trapped between two sequences, both converging
to zero. By the Sandwich Theorem, we can conclude that
lim an = 0
n→∞
as well.
2