MAS221 Analysis, Semester 2 Solutions
Sarah Whitehouse
Chapter 1 Problems: Series
1. Let
sn = a1 + · · · + an .
Since the series
L ∈ R.
P∞
n=1
an converges, we have limn→∞ sn = L, for some
As (sn+1 ) is a subsequence of (sn ), it follows that limn→∞ sn+1 = L. By
the algebra of limits,
lim an+1 = lim sn+1 − sn = lim sn+1 − lim sn = L − L = 0.
n→∞
n→∞
n→∞
n→∞
Hence (an ) converges with limit 0.
2. Write
ak =
B
C
A
+
+
k
k+1 k+2
Then
A(k + 1)(k + 2)Bk(k + 2) + Ck(k + 1) = 1
Set k = 0. Then 2A = 1, so A = 1/2.
Set k = −1. Then −B = 1, so B = −1.
Set k = −2. Then 2C = 1, so C = 1/2.
Thus
ak =
1
1
1
−
+
2k k + 1 2(k + 2)
1
Hence
n
X
1 1 1
1
( ·1−1· + · )
2
2 2 3
ak =
k=1
1
+( ·
2
1
+( ·
2
1
+( ·
2
..
.
1
+( ·
2
1
1 1
−1· +
2
3 2
1
1 1
−1· +
3
4 2
1
1 1
−1· +
4
5 2
1
· )
4
1
· )
5
1
· )
6
1
1
1
1
−1·
+ ·
)
n
n+1 2 n+2
Cancelling terms
n
X
ak =
k=1
1
1 1 1 1
1
1
1
1
·1−1· + · + ·
−1·
+ ·
.
2
2 2 2 2 n+1
n+1 2 n+2
That is
n
X
ak =
k=1
1
1
1
−
+
.
4 2(n + 1) 2(n + 2)
Letting n → ∞, we see that
∞
X
ak =
k=1
1
.
4
3. Observe that
n
X
k=1
1
ln 1 +
k
1
1
= ln(1 + 1) + ln(1 + ) + · · · + ln(1 + )
2
n
1
1
1
= ln (1 + 1)(1 + )(1 + ) · · · (1 + ) .
2
3
n
Now putting everything over a common denominator
1
1
1
2 3 4 5
n+1
(1 + 1)(1 + )(1 + ) · · · (1 + ) = · · · · · · ·
= n + 1.
2
3
n
1 2 3 4
n
So
n
X
k=1
1
ln 1 +
k
= ln(n + 1)
Certainly ln(n + 1) does not converge as n → ∞.
2
4. Since we have assumed an 6= 0, we can consider the sequence ( abnn ). Note
that
bn
lim
= 1,
n→∞ an
using the algebra of limits.
Since limn→∞
bn
an
= 1, we have
|bn | ≤ 2|an |,
for all sufficiently large n. (Taking = 1, there is some N such that for
n > N , | abnn − 1| < 1. Thus for n > N , |bn − an | < |an |, and using the
triangle inequality, |bn | = |bn − an + an | ≤ |bn − an | + |an | < 2|an |.)
P∞
P∞
By assumption
absolutely, that is,
n=1 an converges
n=1 |an | conP∞
verges. Therefore, so does n=1 2|an |.
Hence, by the comparison test, the series
∞
X
|bn |
n=1
converges, that is
P∞
n=1 bn
converges absolutely.
5. (a) Note that
1
1
1
= −
.
n(n + 1)
n n+1
So we have
n
X
k=1
1
1 1 1
1
1
1
= 1 − + − + ··· + −
=1−
.
n(n + 1)
2 2 3
n n+1
n+1
If we let n → ∞, we see
∞
X
k=1
1
= 1.
k(k + 1)
In particular, it converges.
(b) Observe
n2
1
1/n(n + 1)
=
lim
= lim
= 1.
2
2
n→∞
n→∞ n + n
n→∞ 1 + 1/n
1/n
P∞
So by the previous question, n=1 n12 converges.
lim
(c) Observe that nα ≥ n2 , so
1
1
≤ 2.
α
n
n
So the series
P∞
1
n=1 nα
converges by the comparison test.
3
6. (a) Let
an =
1 + (4/3)n
3n + 4n
=
3n + 5n
1 + (5/3)n
Let
n
4
bn =
5
P∞
Then, as a geometic series, n=1 bn converges.
But
bn
1 + (3/5)n
(1/5)n + (1/3)n
=
→1
=
n
n
an
(1/4) + (1/3)
1 + (3/4)n
as n → ∞.
P∞
So by question 4, the series n=1 an converges.
Alternatively, we could use the ratio test, but the arithmetic is harder.
(b) Let
n n
3
5
+
4
4
P∞
→ ∞ as n → ∞. So the series
n=1 an does not
3n + 5 n
an =
=
4n
Note that an
converge.
(c) We know that 1/n → 0 as n → ∞, and cos is a continuous function,
with cos 0 = 1, so
1
=1
lim cos
n→∞
n
It follows that the series
∞
X
cos
n=1
1
n
does not converge.
(d) Let
√
an =
=
Now
n2 + 1 −
n
√
n2 − 1
=
(n2 + 1) − (n2 − 1)
√
√
n( n2 + 1 + n2 − 1)
2
√
√
.
2
n( n + 1 + n2 − 1)
p
√
n2 + 1 ≥ n2 = n,
so
0 ≤ an ≤
p
n2 − 1 ≥ 0,
2
.
n2
P∞
The seriesP n=1 1/n2 converges, so, by the comparison test, so does
∞
the series n=1 an .
4
(e) Let
an =
(n!)2
(2n)!
Then
(n + 1)2
(2n)!((n + 1)!)2
(1 + 1/n)2
an+1
=
=
=
2
an
(2n + 2)!(n!)
(2n + 1)(2n + 2)
(2 + 1/n)(2 + 2/n)
so
an+1
1
= .
an
4
P∞
Hence by the ratio test the series n=0 an converges.
lim
n→∞
7. The n-th partial sum is sn = 1 + 2r + 3r2 + · · · + nrn−1 . Then
(1 − r)sn = 1 + r + r2 + · · · + rn−1 − nrn =
Thus
sn =
1 − rn
− nrn .
1−r
1 − rn
nrn
.
−
2
(1 − r)
1−r
1
Since |r| < 1, limn→∞ rn = 0. So limn→∞ sn = (1−r)
2 , as required.
1
1
8. Following the hint, we let a = tan−1 2n−1
and b = tan−1 2n+1
and
use the identity
tan a − tan b
tan(a − b) =
.
1 + tan a tan b
Then
1
− 1
1
tan(a − b) = 2n−1 12n+1 = 2 .
2n
1 + (2n−1)(2n+1)
Thus
tan−1
1
2n2
= a − b = tan−1
1
2n − 1
− tan−1
1
2n + 1
.
So
1
1
1
1
− tan−1 + tan−1 − tan−1 + · · ·
1
3
3
5
1
1
−1
−1
+ tan
− tan
2n − 1
2n + 1
1
1
π
1
= tan−1 − tan−1
= − tan−1
.
1
2n + 1
4
2n + 1
1
Since tan−1 is continuous at 0, limn→∞ tan−1 2n+1
= tan−1 0 = 0. So
limn→∞ sn = π4 , as required.
sn = tan−1
5
9. (a) False. For example, the harmonic series
but limn→∞ n1 = 0.
P∞
does not converge,
(b) True. The partial sums converge to 0, so
definition.
P∞
an converges to 0, by
1
n=1 n
n=1
(c) True. As the terms an are positive, the sequence of partial sums is
monotonic
increasing. Since it is also bounded above, it converges,
P∞
so n=1 an converges.
P∞
(d) True. Since
n=1 |an | converges, we have limn→∞ |an | = 0. In
particular, there is some N such that for all n > N , |an | < 1. Then,
for n > N ,
a2n = |an ||an | < |an |.
P∞
It follows by the comparison test that n=1 a2n converges.
6