HW6
SOLUTIONS TO STARRED PROBLEMS
3.1.3:
First note that
1
1
=
.
−1
(2n − 1)(2n + 1)
4n2
Using the method of partial fractions, we check that
1
1
1
1
=
−
.
4n2 − 1
2 2n − 1 2n + 1
(Alternatively, we can simplify the right-hand side by getting a common denomi∞
P
1
nator.) Now let {sn } denote the sequence of partial sums of
4n2 −1 . It follows
n=1
that
n
X
1
1
1
sn =
−
2 2n − 1 2n + 1
j=1
1 1 1
1
1 1 1
1
1
=
−
−
−
+
+ ··· +
2 1 3
2 3 5
2 2n − 1 2n + 1
1
1
= −
.
2 2(2n + 1)
Clearly sn →
1
2
as n → ∞, and hence
∞
P
n=1
3.1.4:
∞
P
(a) Note that
n=1
4 n
5
1
4n2 −1
= 12 .
is a convergent geometric series and that, by applying
the method of partial fractions,
∞
P
4
n(n+2)
is a convergent telescoping series. By
∞ P
4
4 n
the Series Combination Rules, it follows that
+ n(n+2)
converges.
5
n=1
n n o
∞
n P
(b) Since the sequence 1 + 21
converges to 1, the series
1 + 12
fails
n=1
n=1
the Non-null Test; i.e. it diverges.
∞ P
√1 − √ 1
(c) It is clear that
is a telescoping series, and hence converges.
n
n+1
n=1
Extra Problem #1:
We apply the method of partial fractions to write
2
1
2
1
= −
+
n(n + 1)(n + 2)
n n+1 n+2
1
1
1
1
=
−
−
−
.
n n+1
n+1 n+2
1
2
HW6 SOLUTIONS TO STARRED PROBLEMS
Let {sn } denote the sequence of partial sums of
∞
P
n=1
2
n(n+1)(n+2) .
Then
n
X
2
j(j
+
1)(j
+ 2)
j=1
1
1 1
1 1
1 1
=
1−
−
−
+
−
−
−
2
2 3
2 3
3 4
1
1
1
1
+ ··· +
−
−
−
n n+1
n+1 n+2
1
1
1
= 1−
−
−
.
2
n+1 n+2
sn =
Clearly sn →
that
1
2
as n → ∞. It follows from this and the Series Combination Rules
∞
X
1
1
= .
n(n + 1)(n + 2)
4
n=1
Extra Problem #2:
Let N ∈ N be such that an = bn for all n ≥ N . Let {sn } be the sequence of
∞
P
partial sums for the series
an and let {tn } be the sequence of partial sums for
the series
∞
P
n=1
bn . By assumption, {sn } converges. On the other hand, since an = bn
n=1
for all n ≥ N , it follows that
sn − tn = sN − tN
for all n ≥ N .
In other words, the sequence {sn −tn } is eventually constant, and hence convergent.
Since tn = sn − (sn − tn ) for all n ∈ N, we see that {tn } is the difference of two
∞
P
convergent sequences. Therefore {tn } converges; i.e.
bn converges.
n=1