MATH 101 V2A
March 4th – Practice problems
Hints and Solutions
1. Which of the following series converge? Justify your claims.
n
∞
X
1
1010
(a)
10
1 − 1010
10
n=1
10
Solution: This series converges as it a geometric series with a = 1010
(b)
and r = 1 −
1
< 1.
101010
∞ n
X
4
n=0
3
Solution: This series diverges as it is a geometric series with a = 1 and r =
(c) 1 − 5 −
9
+2+π+
5
4
3
> 1.
12 13 14
2
2
2
+
+
+ ···
3
3
3
Solution: This series converges. To see this, notice that we can rewrite the series as
1−5−
9
+2+π+
5
12 13 14
12 X
∞ n
2
2
2
9
2
2
+
+
+ ··· = 1 − 5 − + 2 + π +
.
3
3
3
5
3
3
n=0
From this we can see that the series is just the sum of a convergent geometric series (with
12
2
2
9
a=
and r = < 1) and a number (1 − 5 − + 2 − π), and so the series must converge.
3
3
5
∞
X
1
(0.5)n−1
(d)
n
n=1
1
Solution: This series converges by the Comparison Test since for n ≥ 1, (0.5)n−1 ≤ (0.5)n−1 , and
n
P∞
n−1
is a convergent geometric series (with a = 1 and r = 0.5).
n=1 (0.5)
2. Prove that (possibly infinite) decimal expressions represent real numbers; that is, prove that
bk bk−1 · · · b1 b0 . a1 a2 a3 · · ·
is a convergent series.
Hint: The decimal expansion
bk bk−1 · · · b1 b0 . a1 a2 a3 · · · ,
represents the number given by the series
bk · 10k + bk−1 · 10k−1 + · · · + b1 · 101 + b0 · 100 +
∞
X
an ·
n=1
1
10
n
.
To show that the series converges, use the Comparison Test to show that
n
∞
X
1
an ·
10
n=1
converges. (Hint: how large can an be?)
3. The Comparison Test states that if
∞
X
an and
n=1
∞
X
bn are series whose terms are (eventually) all
n=1
non-negative and an ≤ bn (eventually), then
(i) if
(ii) if
∞
X
n=1
∞
X
n=1
bn converges, then
an diverges, then
∞
X
an converges.
n=1
∞
X
bn diverges.
n=1
(a) Prove (i) by showing that the sequence of partial sums sn =
∞
X
an is (eventually) monotone
n=1
increasing and bounded and applying the Monotone Convergence Theorem.
Solution: We will show that (i) is true when an , bn ≥ 0 for all n and an < bn for all n. The more
general case is a consequence of this case since this will show that
∞
X
an
n=N
converges for some N , and the first N − 1 terms of the series cannot affect its convergence.
Since an ≥ 0 for all n, s= sn−1 + an ≥ sn−1 for all n > 1 which means that the sequence of partial
sums is monotone. Now, since an ≤ bn for all n, and bn ≥ 0 for all n,
sN =
N
X
an ≤
n=1
N
X
n=1
bn ≤
∞
X
bn .
n=1
P∞
P∞
Now, if n=1 bn converges, then n=1 bn is just a finite number and therefore
P∞provides an upper
bound for the sequence of partial sums. It now follows from the MCT that n=1 an converges.
2
(b) Prove (ii) by first showing that, since
∞
X
an diverges and an is eventually non-negative,
n=1
must diverge to infinity. Use this and the definition of divergence to infinity to show that
∞
X
n=1
∞
X
an
bn
n=1
also diverges.
Solution: Again, we will show that (ii) is true when an , bn ≥ 0 for all n, and an ≤ bn for all n.
P∞
Since an ≥ 0, n=1 an diverges to infinity. This means that, for any number M > 0, there is a
number N such that sn > M for all n > N . However, since bn > an for all n,
n
X
bk ≥
k=1
for all n > N . That is, the partial sums of
must also diverge to infinity.
n
X
ak = sn > M
k=1
P∞
n=1 bn
3
can be made arbitrarily large, and so
P∞
n=1 bn