EXERCISES FOR CHAPTER 3: Infinite Series
SET 3.1
n
4
4
in partial fractions and hence find the sum .
k(k + 2)
k =1 k(k + 2)
n
4
Determine if lim exists.
n
k =1 k(k + 2)
Solution
A
B
4
= +
. By the cover-up method A = 2 and B = 2 . Hence
k(k + 2) k k + 2
n
2 2
2 4
2 2 2 2
2
k(k + 2) = 1 3 + 2 4 + + n 1 n + 1 + n n + 2 k =1
1. Express
2
2
n +1 n + 2
2n + 3
= 3
(n + 1)(n + 2)
n(3n + 5)
=
(n + 1)(n + 2)
= 2 +1
Thus,
n(3n + 5)
4
k(k + 2) = lim (n + 1)(n + 2) = 3 .
k =1
2. Show that
n
5
5
5
+
+
+ exists and find its value.
1 2 2 3 3 4
Solution
A
B
5
= +
. By the cover-up method A = 5 and B = 5 . Hence
k(k + 1) k k + 1
n
5 5
5 5
5 5 5 5
5
k(k + 1) = 1 2 + 2 3 + + n 1 n + n n + 1
k =1
5
=5
n +1
5n
=
n +1
5
5n
= lim
= 5.
Thus, n
n +1
k =1 k(k + 1)
1
1
1
1
+
+
++
+ .
3. Find the value of the infinite sum
n(n + 3)
1 4 2 5 3 6
Solution
A
B
1
1
1
= +
. By the cover-up method A = and B = . Hence
n(n + 3) n n + 3
3
3
25
K. A. Tsokos: Series and D.E.
N
11
1
1
1 1
1
1 1
1
1 1
1
n(n + 3) = 3 1 4 + 3 2 5 + 3 3 6 + 3 4 7 +
n =1
1 1
1 1 1
1 1 1
1 + + + 3 N 2 N + 1 3 N 1 N + 2 3 N N + 3 1
1 1
1
1
1 = 1 + + 3
2 3 N + 1 N + 2 N + 3
=
N(11N 2 + 48N + 49)
18(N + 1)(N + 2)(N + 3)
11
1
N(11N 2 + 48N + 49)
=
lim
n(n + 3) N 18(N + 1)(N + 2)(N + 3) = 18 .
n =1
Thus
4. Evaluate
1
1
1
+
++
+ .
n(n + 2)(n + 4)
1 3 5 2 4 6
Solution
A
B
C
1
1
1
= +
+
. By the cover-up method A = , B = and
n(n + 2)(n + 4) n n + 2 n + 4
8
4
1
C = . We then write
8
1
2
1
1
=
+
n(n + 2)(n + 4) 8n 8(n + 2) 8(n + 4)
1
1
1
1
=
+
8n 8(n + 2) 8(n + 4) 8(n + 2)
=
11
1 1 1
1 + 8 n (n + 2) 8 (n + 4) (n + 2) so that
1 N 1
1 1 N 1
1 1
=
n(n + 2)(n + 4) 8 n (n + 2) 8 (n + 2) (n + 4) n =1
1
1
N
=
1
1
1 1
1 1
1
1
1
1 )+( ) +
(1 ) + ( ) + ( ) + + (
8
3
2 4
3 5
N 1 N +1
N N +2 1 1 1
1 1
1 1
1
1
1
1 ( ) + ( ) + ( ) ++ (
)+(
)
8 3 5
4 6
5 7
N +1 N + 3
N +2 N +4 =
1
1
1
1 1 + +
8
2 N +1 N + 2
11 1
1
1 + 8 3 4 N + 3 N + 4
=
N(N + 5)(11N 2 + 55N + 62)
96(N + 1)(N + 2)(N + 3)(N + 4)
26
K. A. Tsokos: Series and D.E.
11
1
N(N + 5)(11N 2 + 55N + 62)
= lim
=
Thus, .
N
96(N + 1)(N + 2)(N + 3)(N + 4) 96
n =1 n(n + 2)(n + 4)
5. Find
3
n +1
2
+
++
+ .
(2n 1)(2n + 1)(2n + 3)
1 3 5 3 5 7
Solution
A
B
C
n +1
=
+
+
. By the cover-up method,
(2n + 1)(2n 1)(2n + 3) 2n + 1 2n 1 2n + 3
1
3
1
and C = . Hence we may write:
A= , B=
8
16
16
2 1
3 1
1 1
n +1
=
+
16 2n + 1 16 2n 1 16 2n + 3
(2n + 1)(2n 1)(2n + 3)
=
2 1
1 1 1
1 + 16 2n 1 2n + 1 16 2n 1 2n + 3 Then,
n +1
N
2
1
1
1
1
1
1
1
(2n 1)(2n + 1)(2n + 3) = 16 (1 3 ) + ( 3 5 ) + ( 5 7 ) + + ( 2N 1 2N + 1)
n =1
+
1
1
1 1
1 1
1
1
1
1
)+(
)
(1 ) + ( ) + ( ) + + (
16
5
3 7
5 9
2N 3 2N + 1
2N 1 2N + 3 n +1
N
2
1
1
1
1
1
(2n 1)(2n + 1)(2n + 3) = 16 1 2N + 1 + 16 1 + 3 2N + 1 2N + 3 )
n =1
N(5N + 7)
6(2N + 3)(2N + 1)
5
n +1
N(5N + 7)
= lim
=
Hence, .
N 6(2N + 3)(2N + 1)
24
n =1 (2n 1)(2n + 1)(2n + 3)
=
n
k+3
k+3
in
partial
fractions
and
hence
find
the
sum
.
2
2
k(k 1)
k = 2 k(k 1)
n
k+3
Determine if lim exists.
2
n
k = 2 k(k 1)
Solution
k+3
A
B
C
k+3
=
= +
+
. By the cover-up method, A = 3 ,
2
k(k 1) k(k 1)(k + 1) k k 1 k + 1
B = 2 and C = 1 . Hence we may write:
6. Express
3
2
1
1 1
1 k+3
1
= +
+
= 2
2
k 1 k k k + 1 k k 1 k +1
k(k 1)
so that
27
K. A. Tsokos: Series and D.E.
N
1
1 1
1
1 k+3
= 2 (1 ) + ( ) + + (
)
2
2
2 3
N 1 N 1)
k(k
2
1 1
1
1 1 1
( ) + ( ) ++ ( 2 3
3 4
N N + 1
1 1
1 = 2 1 N 2 N + 1
N 1
2(N 1)
=
2(N + 1)
N
(N 1)(3N + 4)
=
2N(N + 1)
k+3
(N 1)(3N + 4) 3
= lim
= .
2
N
2
1)
2N(N + 1)
and hence
k(k
2
7. Investigate, by using partial fractions, whether the integrals (a)
x
5
and (b)
x
5
2
2
1
dx
5x + 6
x +1
dx exist.
5x + 6
Solution
1
A
B
1
=
=
+
(a) 2
. By the cover up method, A = 1 and
x 5x + 6 (x 3)(x 2) x 3 x 2
1
1
1
B = 1 , so that I = lim 2
dx = lim (
) dx . Then,
x3 x2
x 5x + 6
5
5
I = lim (ln x 3 ln x 2 ) 5
x3
= lim ln
x2 5
3
2
= lim ln
ln
2
3
3
= 0 + ln
2
3
3
= ln lim
= ln1 = 0 .)
2
2
x +1
A
B
x +1
=
=
+
(b) 2
. By the cover up method, A = 4
x 5x + 6 (x 3)(x 2) x 3 x 2
The integral converges. (Notice that lim ln
4
3
x +1
and B = 3 , so that I = lim 2
dx = lim (
) dx . Then,
x3 x2
x 5x + 6
5
5
28
K. A. Tsokos: Series and D.E.
I = lim (4 ln x 3 3ln x 2 ) 5
= lim (4 ln 3 3ln 2 ) (4 ln 2 3ln 3)
16
27
To evaluate the limit, lim (4 ln 3 3ln 2 ) write
= lim (4 ln 3 3ln 2 ) ln
lim (4 ln 3 3ln 2 ) = lim (4 ln 3 4 ln 2 ) + lim ln 2
= 4 lim ln
3
+ lim ln 2
2 = 0 + lim ln 2
=
The integral diverges. (Again, lim 4 ln
3
3
= 4 ln lim
= 4 ln1 = 0 .)
2
2
SET 3.2
Apply the divergence test to the series below to determine those that diverge. If the test
is inconclusive you should say so.
n +1
n2
1. (a) (b) 2
n
1
1 n +1
Solution
n +1
(a) lim
= 1 , so series diverges.
n n
n2 + 1
(b) lim 2 = 1 , so series diverges.
n n
1
n
(c) ) lim
= , so series diverges.
n 2n + 5
2
2. (a)
(1+ (1) )
n
(c)
1
(b)
1
(e)
n
2n + 5
(c)
( ln(n + 2) ln n )
2
1
Solution
(a) lim 1 + (1)n does not exist so series diverges.
n
(
)
(b) lim un = e 0 , so series diverges.
n
n + 2
n + 2
= ln lim (c) lim ( ln(n + 2) ln n ) = lim ln = ln1 = 0 , inconclusive.
n
n
n n n 3. (a)
e
1
n
(b)
e
1
n
(c)
ne
n
1
Solution
(a) lim e n = 0 , so test is inconclusive. (Series is geometric and converges.)
n
29
K. A. Tsokos: Series and D.E.
(b) lim en = , so series diverges.
n
(c) lim ne n = 0 , so test in inconclusive.
n
4. (a)
(
n + n n)
2
1
Solution
(a) lim
n
(
)
n2 + n n =
1
(b) (c)
n +1 n 1
1 2 1
n
1
by the example in the text so series diverges.
2
(b)
1
1
= lim
n
n +1 n
n +1 n
lim
n
n +1 + n
n +1 + n
n +1 + n
n
n +1 n
= lim n + 1 + n
= lim
n
=
so series diverges.
n
1 (c) lim = 0 , so test is inconclusive. (Series is geometric and converges.)
n 2
(1)n
e n 1
)
(b) n n
(c) n
5. (a) (1 n
1
1
1
(1)n (a) lim 1 = 1 , so series diverges.
n n (b) lim n n = 1 , so series diverges.
n
en 1
(c) lim n = 0 , so test is inconclusive. (Series is geometric and converges.)
n n 2 n
n + 3 n
6. (a) (b) n n 1
1
Solution
n
n 2
2
(a) lim = e , so series diverges.
n n n n
(c) n + 3
1
n + 3
3
(b) lim = e , so series diverges.
n n n
n
n 3
(c) lim = e , so series diverges.
n n + 3 7. (a)
1
ln n
2
Solution
(b)
n + 1
n ln
1
(c)
n ln n + 1
1
30
K. A. Tsokos: Series and D.E.
1
= 0 , so test is inconclusive.
n ln n
n +1
(b) lim ln(
) = 0 , so test is inconclusive.
n
n
n
(c) lim ln(
) = 0 , so test is inconclusive.
n
n +1
(a) lim
1
1
cos
(b)
n n sin n 1
1
Solution
1
(a) lim cos = 1 , so series diverges.
n
n
1
sin
1
n = 1 , so series diverges.
(b) lim n sin = lim
n
n
1
n
n
n
(c) lim
= 1 , so series diverges.
n n + 2
8. (a)
9. (a)
n!
5
(b)
n
n=1
nn
n!
n=1
(c)
n
n+2
1
(c)
2n!+7
10
+ n!
n
1
Solution
n!
(a) lim n = , so series diverges.
n 5
nn
(b) lim = , so series diverges.
n n!
2n!+ 7
(c) lim 10
= 2 , so series diverges.
n n
+ n!
2n + 7 10. 3n 2 1
Solution
n
2n + 7 lim lim
= n
n 3n 2 n
7 n
n
)
7 /2
2n = 2 e = 0 , test is inconclusive.
2 / 3
2
3 e
(3n)n (1 )n
3n
(2n)n (1 +
SET 3.3
Apply, if possible, the integral test on the following series. If the test cannot be applied,
state why.
31
K. A. Tsokos: Series and D.E.
1. (a)
n
n +1
1
(b)
n
n2 + 1
1
(c)
n4
4n5 + 2
1
Solution
(a) Terms are increasing because
d n (n + 1) n
1
=
> 0 , test cannot be
=
2
(n + 1)
dn n + 1
(n + 1)2
applied.
(b) Terms are decreasing because
1 n2
d n (n 2 + 1) n 2n
=
=
< 0 and
(n 2 + 1)2
dn n 2 + 1 (n 2 + 1)2
n
n
1
lim 2
dn = lim ln(n 2 + 1) 1 = so series diverges.
1 n 2 + 1 dn = n +1
2 1
(c)
4n 3 (4n 5 + 2) n 4 20n 4 8n 3 4n 8 4n 3 (2 n 5 )
d
n4
=
=
=
< 0 for n > 1 so
(4n 5 + 2)2
(4n 5 + 2)2
dn 4n 5 + 2
(4n 5 + 2)2
terms are decreasing. lim 1
1
1
n4
=
lim ln(4 5 + 2) ln 6 = , so series
5
20
4n + 2 20 diverges.
1
2. (a) n ln n
2
ln n
(b) 2
n
1
(c)
1
ln n
n
Solution
(a) Terms are clearly decreasing. The integral is lim 1
1
dn . Make the substitution
n ln n
dn
so that
u = ln n du =
n
ln 1
1
ln lim
dn = lim du = ln u ln 2 = so series diverges.
n ln n
u
2
ln 2
1 2
d ln n n n 2n ln n 1 2 ln n
=
=
< 0 so terms are decreasing. The integral is
(b)
n3
dn n 2
n4
ln 2 1
ln n
1
ln n ln n 1
lim 2 dn = lim + lim 2 dn = lim lim =
+
n
n
n
n
2
2
n
2
2
2
2
2
and so series converges.
n
1
ln n
2 ln n
d ln n
ln n
n
2
n
2
=
=
< 0 for n > e . The integral is lim (c)
dn .
3/2
2n
n
dn n
n
1
Using integration by parts,
32
K. A. Tsokos: Series and D.E.
ln n
n
lim dn = 2 lim n ln n 2 lim dn = 2 lim n ln n 4 lim n
1
1
1
n
n
1
1
= lim 2 n(2 + ln n) = 1
and the series diverges.
en
3. (a) n
(e + 1) 2
1
(b)
ne
n
(c)
1
1
2
n
1
Solution
en (en + 1)2 en 2(en + 1)en en (1 en )
d
en
=
= n
< 0 so terms are decreasing.
(a)
(en + 1)4
dn (en + 1)2
(e + 1)3
en
en
dn
=
lim
dn
(e n + 1)2
(e n + 1)2
1
1
lim 1
= lim n
e + 1
1
1
=
e +1
so series converges.
d n en nen en (1 n)
ne =
=
< 0 for n > 1 so terms are decreasing.
(b)
e2n
dn
e2n
lim ne dn = lim ne
n
n 1
1
+ lim e n dn = lim ne n 1 lim e n 1
1
= 1 + e1
so series converges.
(c) The terms are obviously decreasing. The integral gives
2 n
1
and so series converges.
lim 2 dn = lim
=
ln 2
2
ln
2
1
1
n
1
4. (a) 2n + 1
1
1
(b) n+5
1
(c)
ne
n 2
1
Solution
(a) The terms are obviously decreasing. The integral gives
1
1
lim dn = lim ln(2n + 1) 1 = and so series diverges.
2n + 1
2 1
(b) The terms are obviously decreasing. The integral gives
1
lim dn = lim ln(n + 5) 1 = and so series diverges.
n + 5
1
33
K. A. Tsokos: Series and D.E.
2
2
d n2 en nen (2n) 1 2n 2
ne =
=
< 0 and so terms are decreasing.
(c)
2
2
dn
en
e2n
2
2 1
1
lim ne n dn = lim e n = so series converges.
1
2 2
1
ln n
5. (a) n
2
en
(c) n
+1
1 e
n
(b) ln n
2
Solution
d ln n 1 ln n
(a)
=
< 0 for n > 2 so terms are decreasing. Making the substitution
n2
dn n
ln dn
ln n
(ln n)2
the integral is lim u = ln n du =
dn = lim u du = lim
) = and
n
2
n
2
ln 2
2
the series diverges.
ln n 1
d n
=
> 0 for n > 2 so test cannot be applied.
(b)
dn ln n (ln n)2
(c) The general term can be re-written as
1
en e n and is obviously
=
n n
e e + 1 1 + en
decreasing. The integral is
e n
lim
dn = lim ln(e n + 1) 1 = lim ln(e + 1) + ln(e1 + 1) = ln(e1 + 1) and so
e n + 1
1
series converges.
1
(b) (c)
1 (n + 1)(n + 2)
1
6. (a) n
e +1
1
n
1
1
+1
2
Solution
(a) Terms are clearly decreasing.
1
e n
lim n
dn = lim n
dn = lim ln(e n + 1) 1 = lim ln(e + 1) + ln(e1 + 1)
e + 1
e
+1
1
1
= ln(e1 + 1)
and so series converges.
(b) Terms are clearly decreasing.
1
1
1
lim dn = lim 2
dn = lim dn
(n + 1)(n + 2)
n + 2n + 2
(n + 1)2 + 1
1
1
1
arctan 2
2
and so series converges.
(c) Terms are clearly decreasing.
1
lim 2
dn = lim arctan (n) 1 = arctan1 =
n + 1
2
4
1
and so series converges.
= lim arctan (n + 1) 1 =
34
K. A. Tsokos: Series and D.E.
7. (a)
n
1
1
2
+ 4n + 5
(b)
sin
1
n
2
(c)
n2
n4 + 1
1
Solution
(a) Terms are clearly decreasing.
1
1
lim 2
dn = lim dn = lim arctan (n + 2) 1 = arctan 3
2
n + 4n + 5
(n + 2) + 1
2
1
1
and so series converges.
(b) Test cannot be applied. The terms are not positive and are not decreasing.
2n(n 4 + 1) n 2 4n 3 2n(1 n 4 )
d n2
=
=
< 0 so terms are decreasing. The
(c)
(n 4 + 1)2
dn n 4 + 1
(n 4 + 1)2
integral is lim 1
n2
n2
1
1
dn
<
lim
dn = lim 2 dn = lim
= 1 and so the series
4
4
n
n
n
n +1
1
1
1
converges.
8. (a)
1
1
1+
(b)
n
n
2
(c)
n
1
n
3
1
1
+1
Solution
1
dn . Make the substitution
1 +
n
1
(a) Terms are clearly decreasing. The integral is lim u = 1 + n du =
1
2 n
dn . Then
1
u 1
lim
dn = 2 lim du = 2 lim ( 2) ln = , and so series diverges.
1 +
2
u
n
1
2
(b)
n2 n
n2 n
2 n
1
1
lim n2 dn = lim
+
lim 2 n dn = lim
+
lim
ln 2
ln 2
ln 2 1
ln 2 ln 2 1
1
1
1
n
1 + ln 2
1
1
lim lim +
2
ln 2 2
(ln 2) 2
2(ln 2)2
Both limits are clearly zero and so series converges.
3n 2
d 1
=
< 0 so the terms are decreasing. The integral is
(c)
(n 3 + 1)2
dn n 3 + 1
=
lim 1
1
1
1
1
dn < lim 3 dn = lim 2
3
n
2 n
n +1
1
=
1
1
and so the series converges.
2
9. For what values of p does the following series converge:
2
Solution
1
n(ln n)
p
?
35
K. A. Tsokos: Series and D.E.
dn
dn
dn
Terms are obviously decreasing. . Call u = ln n du =
= lim p
p
n(ln n)
n
n(ln n)
2
2
ln ln dn
du
u1 p
(ln )1 p ln 21 p
and so the integral is .
= lim
= lim
= lim
n(ln n) p ln2 u p 1 p ln 2 1 p
1 p
2
The limit is zero if p > 1 and infinite if p < 1 . If p = 1 the integral must be redone to
give
ln dn
du
ln =
lim
lim ln u ln 2 = lim (ln ln ) ln ln 2 .
2 n(ln n) ln2 u = Hence the series converges for p > 1 and diverges for p 1 .
SET 3.4
Apply the ratio test to these series. If the ratio test is inconclusive apply a different test.
1. (a)
5
3n
1
(b)
5n
3n
1
(c)
en
n
1
Solution
5
n+1
1
(a) = lim 3 = < 1 so series converges.
n 5
3
n
3
5 n+1
n+1
5
(b) = lim 3 n = > 1 so series diverges.
n 5
3
n
3
en+1
n+1
e
(c) = lim n = < 1 so series converges.
n e
n
n + 1 n
1 n 2. (a)
(b)
7n
4n + 1
1
(c)
e
n
n!
1
Solution
(a) = lim
n + 2
n +1
n +1
= lim
1 1 +
n + 1
n
1
n + 1
1 + n
n
diverges by the divergence test.)
n
n
n
n +1
=
e
= 1 so test is inconclusive. (Series
e
36
K. A. Tsokos: Series and D.E.
7 n +1
4 n (1 +
n +1
n
n +1
7
4
+
1
(b) = lim 4 n+ 1 = lim n n +1
= 7 lim
n
n 7
n n +1
7
4 +1
4 (1 +
n
4 +1
(n + 1)!
n +1
(n + 1)
= lim
= , series diverges.
(c) = lim e
n
n
n!
e
en
n
2
1+ n 2
3. (a) (b) 2
n
1 n!
1
1
)
4 n = 7 > 1 , series diverges.
1
) 4
n +1
4
2n
(c) n
1 3 +1
Solution
2 n +1
2
(n + 1)!
(a) = lim
= lim
= 0 , series converges.
n
n
n n + 1
2
n!
(b)
1 + (n + 1)2
1 2
2
2
2 1+
2
(n + 1)
(n + 1)2 n 1 + (n + 1) n (n + 1) =
lim
= lim
=
lim
=1
n
n n + 1 1 + n2
n2 1 + 1 1 + n 2 n n + 1 n2 n2
so test is inconclusive. (Series diverges by the divergence test.)
2 n +1
1
3n (1 + n )
n +1
n
n +1
2
2
3
+
1
3
(c) = lim 3 n+ 1 = lim n n +1
== 2 lim
= < 1 , series converges.
n
n 2
n n +1
1
2
3 +1
3 (1 + n +1 ) 3
n
3
3 +1
ln n
ln n
4. (a) (b) (c) nen
n
n
2
2
1
Solution
ln(n + 1)
ln(n + 1) n
(n + 1)
= lim
= 1 by using L’ Hôpital’s rule on each
(a) = lim
n
n
ln n
ln n n + 1
n
fraction. Hence test is inconclusive. (Series diverges by the integral test.)
ln(n + 1)
ln(n + 1)
n
(b) = lim n + 1 = lim
= 1 1 = 1 by using L’ Hôpital’s rule on
n
n
ln n
ln n
n +1
n
each fraction. Hence test is inconclusive. (Series diverges by the integral test.)
n +1
n +1
en n + 1 1
= lim n +1
= < 1 , series converges.
(c) = lim e
n
n e
n
e
n
en
37
K. A. Tsokos: Series and D.E.
1 + n4
(2n)!
5. (a) (b) n
(c) e n ln n
3
n!
2
1
1
Solution
1 + (n + 1)4
1 1 + (n + 1)4
(n + 1)!
(a) = lim
= lim
= 0 1 = 0 < 1 , series converges.
n
n n + 1
1 + n4
1 + n4
n!
(2n + 2)!
n +1
(2n + 2)(2n + 1)
= lim
, series diverges.
(b) = lim 3
n
n
(2n)!
3
3n
ln(n + 1)
n +1
1 ln(n + 1) 1
= lim
= < 1 (used L’ Hôpital’s rule), so series
(c) = lim e
n
n e
ln n
e
ln n
n
e
converges.
2n + 1 n
5n
n
6. (a) n
(b) 2
(c) 5n 3 1
1 3
1 n +1
Solution
5(n + 1)
n +1
3n n + 1 1
= lim n +1
= < 1 , series converges.
(a) = lim 3
n
n 3
5n
3
n
n
3
n +1
n2 + 1 n + 1
n2 + 1
n +1
(n + 1)2 + 1
= lim
=
lim
lim
= 1 by using L’
(b) = lim
2
2
n
n (n + 1) + 1 n
n (n + 1) + 1 n n
n
n2 + 1
Hôpital’s rule, so test is inconclusive. (Series diverges by the integral test.)
(c)
n +1
n +1
2n + 3 2n + 3 2n + 1
5n + 2 5n + 2 = lim
=
lim
n
n +1
n
n
2n + 1 2n + 1 5n 3
5n 3
5n 3 3 n +1
3
) (5n)n +1 (1 )n +1 2n + 1
2n
5n
= lim
n
2
1
(5n)n +1 (1 + )n +1 (2n)n +1 (1 + )n +1 5n 3
5n
2n
3/2 3/5
e e
2
= 2 /5 1/2
e e 5
2
= <1
5
hence series converges.
5 n
2 n+1 n3
7. (a) (b)
1 9n + 1
3n
1
(2n)n +1 (1 +
5n + 3 n
1 2n 1
(c)
38
K. A. Tsokos: Series and D.E.
Solution
2 n + 2 (n + 1)3
3
3
2
2
2
n + 1
n + 1
3n +1
(a) = lim
=
lim
=
lim
= < 1 so series converges.
n +1 3
n
2 n
3 n n 3 n n 3
n
3
(b)
n +1
5 1
n
(1 + )n
9n
+
1
1
1
9n + 10
9n
= lim
= 5 lim = 5 lim
n
n
n 9n + 10 9n + 10
n
10
5 (1 + )n 9n + 10
9n
9n + 1
e1/9
1
= 5 10 /9 lim
= 0 <1
n
e
9n + 10
so series converges.
(c) Working as in 6 (c) the limit is =
5
> 1 so series diverges.
2
2n + 1
3n
n3
(b)
(c)
1 2 n + 1
1 n!
3n
1
Solution
2 n +1 + 1
1
2 n (1 + n )
n +1
n +1
1
1
2
2
+
1
2
(a) = lim 3n
= lim n
= lim
= < 1 , series converges.
n 2 + 1
3 n 2 + 1 3 n 2 n +1 (1 + 1 ) 3
n
2 n +1
3
3n +1
1
2 n (1 + n )
n
n +1
3
2
+
1
2
(b) = lim 2 n+ 1 = 3lim n +1
= 3lim
= > 1 , series diverges.
n
n 2
n n +1
1
3
+1
2 (1 + n +1 ) 2
n
2
2 +1
3
(n + 1)
3
1 n + 1
(n + 1)!
(c) = lim
= lim
= 0 1 = 0 < 1 , series converges.
n
n n + 1 n3
n n!
n
nn
n
n!
9. (a) (b) (c) 2
(2n) n
1 (n!)
1 n!
1
Solution
(n + 1)n +1
n +1
n +1
1
n! n + 1 n (n + 1)!
(a) = lim
= lim
lim
n = n
1 + = 1 e = e > 1 ,
n
n (n + 1)! nn
n
n n + 1
n!
series diverges.
8. (a)
39
K. A. Tsokos: Series and D.E.
(n + 1)!
n +1
n +1
1 (n + 1)! 2n 1
n + 1
(2n + 2)n +1
= lim
= lim
= lim
1 n
n
n!
n + 1
n! 2n + 2 2n n 2n (b)
(2n)n
1
= e1 < 1
2
and the series converges.
(c)
(n + 1)n +1
2
n +1
n +1
1
n! n + 1 n ((n + 1)!)2
= lim
= lim lim
n = n
1 + = 0 e = 0 < 1 ,
n
n (n + 1)!
nn
n n
(n + 1)2 2
(n!)
series converges.
3n + 1
5n + 1
n
10. (a) n
(b) (c) n
n
3
3
n5
1
1
1
n +1
5 +1
1
n +1
5
(1
+
) 1
n +1
n +1
n +1
1
5
5 +1 1
3
5
(a) = lim n
= lim n
= lim
= 5 = >1
n 5 + 1
n
n
1
3
3
3
5 +1 3
5 n (1 + n )
n
5
3
and so the series diverges.
(b)
3n +1 + 1
1
n +1
3
n 3n +1 + 1 1
n 3 (1 + 3n +1 ) 1
(n + 1)5 n +1 1
= lim
= lim
= lim
= 1 3 = < 1,
n
n
n
3 +1
5 n n + 1 3 + 1 5 n n + 1 3n (1 + 1 )
5
5
n
n
3
n5
series converges.
n +1
n +1
1
1
n +1 1
= lim
= 1 = < 1 and the series converges.
(c) = lim 3
n
n
n
3
3
3
n
n
3
SET 3.5
Determine the convergence or divergence of the series by using one of the comparison
tests.
1. (a)
1
1
n(n + 1)
(b)
10
3n n!
1
(c)
2
ln n
2
Solution
(a) Series behaves like the divergent
1
n . Using the limit comparison test we have that
1
40
K. A. Tsokos: Series and D.E.
1
n(n + 1)
n
lim
= lim
= 1 and so the series diverges.
n
n
1
n(n + 1)
n
10
10 10
(b) Since n < n and the series n converges being geometric, the other series
3 n! 3
1 3
converges as well.
1
1
1
2
(c) Since
>
> and the series diverges, the other series diverges as well.
ln n ln n n
2 n
1
2. (a) n
3 +1
1
Solution
(a) 3n + 1 > 3n (b)
ln n 2
(c) n 2
n
(b) 2
1 n +1
1
1
< n . Hence series converges because
n
3 +1 3
n
1
n
< 2 = 3/2 . Hence series converges because
2
n
n +1 n
n
does.
1
1
n
1
3
does.
3/2
1
2
(c) For large enough n, ln n < n
1/8
(ln n) < n
2
1/ 4
n1/ 4
1
ln n < 2 = 3/ 4 . Hence series
n n
n
converges.
5
n4 1
2
Solution
3. (a)
2n 1
3n 1
1
(b)
(c)
1
(a) Use the limit comparison test with the convergent series
2
3n + 1
5
n
4
.
1
5
n4
lim n 1 = lim 4
= 1 so the series converges.
n
n n 1
5
n4
4
2n
(b) Use the direct comparison test with the convergent series n .
1 2.5
n
n
n
2
2
2 1
< n
<
(because 3n 1 > 2.5 n for n > 1) and so the series converges.
n
n
3 1 3 1 2.5
1
2
1
2
(c)
and so series diverges because does.
>
=
3n + 1 3n + n 2n
1 2n
ln n
2n + 3
7
4. (a) 3/2
(b) 2
(c) n
n3
n + n +1
1
2 n
1
Solution
1
1
ln n n1/ 4
(a) 3/2 < 3/2 = 5 / 4 and so series converges since 5 / 4 does.
n
n
n
2 n
7
7
7
(b) 2
< 2 and so series converges since 2 does.
n + n +1 n
1 n
41
K. A. Tsokos: Series and D.E.
(c) Compare with the convergent
1
3
n
using the limit comparison test:
1
2n + 3
n
2n + 2
lim n3 = lim
= 2 and so series converges.
n
n
1
n
3n
1
lnn
n+5
(b) sin 2
(c) 2
5. (a) 2
n
2 n
2 2n 1
1
Solution
1
1
ln n n1/ 4
(a) 2 < 2 = 7 / 4 and so the series converges because 7 / 4 does.
n
n
n
1 n
1
(b) Compare with the convergent series 2 :
1 n
2
1
1
sin
sin
n = lim n =1
lim
n
n 1
1 n2
n and so the series converges.
2
n+5
2
1
n 2 + 5n 1
=
(c) The series behaves as the divergent and since lim 2n 1 = lim 2
n
n 2n 1
1
2
2 n
n
the series diverges.
1
1
n
(b) (c) 6. (a) n
n(n + 1)(n + 2)
1
1 2 1
n3 1
2
Solution
1
(a) Use the limit comparison test with the convergent series n :
1 2
1
n
lim 2 1 = 1 and so the series converges.
n
1
2n
1
(b) Use the limit comparison test with the convergent series 3/2 :
1 n
1
n(n + 1)(n + 2)
n 3/2
lim
= lim
= 1 and so the series converges.
n
n
1
n(n + 1)(n + 2)
n 3/2
1
(c) Use the limit comparison test with the divergent series 1/2 :
1 n
42
K. A. Tsokos: Series and D.E.
n
3
n 3/2
lim n 1 = lim
= 1 and so the series diverges.
n
n
1
n3 1
n1/2
7. (a)
1
n +1
(b)
n2 + 1
lnn
2 n2 + 1
1
ln lnn
(c)
2
Solution
(a) Use the limit comparison test with the divergent series
1
n
1/2
:
1
n +1
2
lim n + 1 = lim
n
n
1
1/2
n
n1/ 4
ln n
< 2 =
(b) 2
n +1 n
n2 + n
n2 + 1
1
n
7/4
= 1 and so the series diverges.
and so series converges since
( )
sin
2
n
1
7/4
does.
1
(c) ln(ln n) < (ln n)1/2 < n1/2
8. (a)
n
1/2
= n1/ 4 . Hence
(b)
21/n
n1/2
1
Solution
(a) Compare with the divergent
1
n : lim
2
1
n
n
(b) Compare with the divergent series
sin
1
n
1/2
1
1
> 1/ 4 and so the series diverges.
ln(ln n) n
1
(c) 1 n!
n = and so the series diverges.
:
1
21/n
1/2
lim n = lim 21/n = 1 and so the series diverges.
n 1
n
1/2
n
1
1
(c)
< 2 for n > 3 and so the series converges.
n! n
1+ cos n
4n
n4
9. (a) n
(b) 4
(c) n
4 + n4
1
1 n
1 4
Solution
1
2
1 + cos n
(a) We have that
.
Compare
the
series
with
<
n
4
n
4
n
4
4 +n
4 +n
1 4 +n
2
n
4
2 4n
2 4n
1+ cos n
.
Hence
converges.
lim 4 + n = lim n
=
lim
=
2
4
n
4
n
n 4 + n 4
n
1
n
4
+
n
n
1
4 (1 + n )
4n
4
1
4
1
n
:
43
K. A. Tsokos: Series and D.E.
4n
4n 4n
(b) Because n < 2 for n > 16 (proof by induction), 4 > n . Since n diverges so
n
2
1 2
n
4
does 4 .
n
1
2n
n4
n4 2n
(c) n < n for n > 16 . Since n converges so does n .
4
4
1 4
1 4
n
1
3
n!
(b) n
(c) n
10. (a) 5
n!
1 2 n!
1
1
Solution
n
n
! 1
3
n
2
n
(a) 3 < ! for n 44 . Hence
<
< 2 and so series converges.
2
n!
n
n!
1
n
1
1
(b) lim 2 n! = 0 . Since converges so does n .
n
1
1 2 n!
1 n!
n!
n!
n!
n
(c) 5 n < ! for n 130 . Hence n >
> n 2 . Hence series diverges.
2
n
5
!
2
2
1
sin n
3
11. (a) n
(b) (c) n!
1 n
1
1 n+ 3
Solution
1
1
1
1
(a) Since n n > n 2 we have that n < 2 . But 2 converges and thus so does n .
n
n
1 n
1 n
2
sin n 1
< , series converges.
(b) Since
n!
n!
1
(c) Using the limit comparison test with the diverging series : we have
1 n
3
lim n + 3 = 3 and so series diverges.
n
1
n
SET 3.6
4
n
Apply any test to determine the convergence or divergence of the following infinite
series.
1. (a)
ne n
2
(b)
1
3k
4k + 1
1
(c)
k!
5
k
1
Solution
2
2
d n2 en nen (2n) 1 2n 2
ne =
=
< 0 and so terms are decreasing.
(a)
2
2
dn
en
e2n
K. A. Tsokos: Series and D.E.
lim ne n dn = 1
2
2
1
lim e n
2 1
=
44
1
so series converges.
2
3k +1
k +1
3
3k +1
4k + 1
(b) By the ratio test, = lim 4 k+ 1 = lim k lim k +1
= < 1 so series converges.
k
k 3
k 4
3
+1 4
k
4 +1
(k + 1)!
k +1
5k
= lim k +1 lim (k + 1) = so series diverges.
(c) By the ratio test, = lim 5
k
k 5
k
k!
k
5
k3
3k + 2 k
1
2. (a) k
(b) 4
(c) 5k
1 e
1 k + k +1
1
Solution
1
k +1
1
(a) By the ratio test, = lim e = < 1 so series converges.
k 1
e
ek
k3
4
1
k4
= 1 so series
(b) Series behaves like the divergent : lim k + k + 1 = lim 4
k
k k + k + 1
1
1 k
k
diverges.
3k +1 + 2 k +1
k +1
5k
3k +1 + 2 k +1 3
5
(c) By the ratio test, = lim k
= lim k +1 lim k
= < 1 so series
k
k 5
k 3 + 2 k
3 + 2k
5
k
5
converges.
1
3k
arctan k
3. (a) (b) k
(c) 3
k
1
1 2 k!
1 k
Solution
arctan k
1
k
= lim arctan k = and so series diverges.
(a) Compare with : lim
k
k
1
2
1 k
k
1
k +1
1
2 (k + 1)! 1
= lim
= 0 < 1 and the series
(b) By the ratio test = lim
k
1
2 k k + 1
2 k k!
converges.
3k +1
3
(k + 1)3
k (c) By the ratio test = lim
= 3 lim = 3 > 1 and the series diverges.
k
k k + 1 3k
k3
45
K. A. Tsokos: Series and D.E.
4. (a)
3
1 + cos k
(b) k
1
2k
k
k
1
k4
(c) k
1 4
Solution
2 k +1
k +1
k + 1 = 2 < 1 and the series converges.
(a) By the ratio test = lim 3
k
k
2
3
k
3 k
1 + cos k 1
> and so the series diverges by the direct comparison test.
(b)
k
k
(k + 1)4
k +1
1
(c) By the ratio test = lim 4 4 = < 1 and the series converges.
k
k
4
k
4
k
1
1
k 2
5. (a) 1
(b) k
(c) 1+
k 1
1 2 +1
1
k k
Solution
1
1+1/
k
1
k
1
= lim 1+1/ k = lim 1 = 1 and so series diverges by
(a) Compare with : lim k
k
k
k
1
k
1 k
kk
k
the limit comparison test.
1
1
1
(b) k
< k and so series converges since k does.
2 +1 2
1 2
k 2
2
(c) lim = e 0 so series diverges by the divergence test.
k k
k
6. (a)
2
1
(b) 2 ln k
1
k 1
2
(c)
1
ln(ln k)
2
Solution
1
(a) Series behaves like
1
k :
2
lim
k
k
k
k 2 1 = lim
= lim
= 1 and so
k
1
1
k 2 1 k
k 1 2
k
k
series diverges.
1
1
> 1/2 and so series diverges.
ln k k
1
1
1
>
> and so series diverges since
(c) ln(ln k) < ln k < k ln(ln k) ln k k
(b) ln k < k 1/2 k
1 7k
Solution
7. (a)
(b)
1
k +1
k +1
(c)
(2k)!
(k!)
2
2
1
k
2
does.
K. A. Tsokos: Series and D.E.
46
k +1
k +1
1
7k
k +1 1
7
= lim k +1 lim
= 1 = < 1 and the series converges.
(a) = lim
k
k 7
k k!
k
7
7
k
7
k +1
1
k+ k
(b) Compare with : lim k + 1 = lim
= 1 and so series diverges since
k k + 1
k k 1
1
k
1
k does.
1
(c) By the ratio test series diverges because:
(2k + 2)!
(2k + 2)!
(k!)2
(2k + 2)(2k + 1)
((k + 1)!)2
= lim
= lim
lim
= lim
= 4 1>1
2
k
k
k
(2k)!
(2k)! k ((k + 1)!)
(k + 1)2
(k!)2
k
1
k!
8. (a) k
(b) (c) k +k
1 e
1 (k + 1)(k + 2)(k + 3)
1
Solution
(a) By the ratio test the series diverges because:
(k + 1)!
k +1
ek
(k + 1)! 1
= lim e
= lim k +1 lim
= =
k
k e
k
k!
e
k!
ek
k
1
k
< 3 = 2 and so the series converges by the direct comparison
(b)
k
(k + 1)(k + 2)(k + 3) k
test.
1
1
k
= 1 and so series diverges since
(c) Compare with : lim k + k = lim
k
k k +
1
k
1 k
k
1
k does.
1
1
4k
k1 0 0
9. (a) (b) (c) k
k
k!
1
1 ln k + k
1 5 +2
Solution
(a) By the ratio test the series converges because:
(k + 1)100
100
k!
(k + 1)!
k + 1
= lim
= lim lim
= 1 0 = 0
100
k
k k
k k (k + 1)!
k!
47
K. A. Tsokos: Series and D.E.
1
1
k
= 1 and so the series
(b) Comparing with the series : = lim ln k + k = lim
k
n
1
ln k + k
1 k
k
diverges.
(c) By the ratio test the series converges because:
4 k +1
k +1
k +1
1 4
4 k +1
5k + 2k
5
+
2
= lim
= lim k lim k +1
= 4 = <1
k
k +1
k
k
k
4
5 5
4
5 +2
k
k
5 +2
1
1
1
10. (a) (b) 2
(c)
2
2
1
1
1 1+
1 1+ 2 ++ k
1 1 + 2 ++ k
++
k
2
Solution
2
k(k + 1) k 2
1
2
> , we have that
< 2 . But 2
2
2
1 + 2 + + k k
1 k
1
converges so by the direct comparison test so does .
1 1+ 2 ++ k
1
1
(b) Since 12 + 2 2 + + k 2 > k 2 , we have that 2
< 2 and so series
2
2
1 + 2 + + k
k
1
converges since 2 converges.
1 k
1
1
(c) From Chapter 2 we know that 1 + + + < 1 + ln k < 1 + k . Hence
2
k
1
1
1
>
. But diverges (using the integral test for example) and
1
1 k +1
1
+
k
1
1 + + +
2
k
thus so does our series.
k
k!2 k
(k + 2)!
1
11. (a) k (b) k
(c) 2
3
k!2 k
1
1
1
(a) Since 1 + 2 + + k =
Solution
(a) Using the ratio test series converges because
k +1
1
(k + 1) 2
1
= lim
=
k
n
2
1
k 2
(k + 1)!2 k +1
2
(k + 1)! 2
3k +1
(b) = lim
= lim
= lim (k + 1) = and so series diverges.
k
n
n
k!2
3
3 n
k!
3k
48
K. A. Tsokos: Series and D.E.
(k + 3)!
1
k!(k + 3)!
(k + 3) 1
2 (k + 1)! 1
= lim
= lim
= < 1 and so series
(c) = lim
n
(k + 2)!
2 n (k + 1)!(k + 2)! 2 n (k + 1) 2
2 k k!
converges.
k +1
12. (a)
1
k +1 k 1
k
(b)
1
k2 + 1 k2 1
k
(k 2 + 1)4
(k 3 + 2)3
1
(c)
Solution
(a) Using the limit comparison test and comparing with the divergent series
1
have that:
k +1 k 1
k
lim
= lim k ( k + 1 k 1)
k
k
1
k
k ( k + 1 k 1)( k + 1 + k 1)
= lim
k
k +1 + k 1
= lim
k
2 k
2 k
= lim
k + 1 + k 1 k k ( 1 + 1 / k + 1 1 / k )
=1
and so the series diverges.
1
k
(b) Compare with the convergent series
2
.
1
lim
k
k2 + 1 k2 1
k
= lim k( k 2 + 1 k 2 1)
k
1
2
k
= lim
k( k 2 + 1 k 2 1)( k 2 + 1 + k 2 1)
k2 + 1 + k2 1
k
= lim
k
2k
k2 + 1 + k2 1
=1
and so series converges.
(c) Compare with the divergent series
1
k .
1
= lim
k
1
k
2k
k( 1 + 1 / k 2 + 1 1 / k 2 )
we
49
K. A. Tsokos: Series and D.E.
(k 2 + 1)4
k(k 2 + 1)4
(k 3 + 2)3
lim
= lim 3
k
k (k + 2)3
1
k
k 9 (1 + 1 / k 2 )4
= lim 9
k k (1 + 2 / k 3 )3
=1
and so the series diverges.
(5k)!
13. (a) k
1 10 (3k)!(2k)!
9 k (3k)!k!
(b) (4k)!
1
Solution
(a) By the ratio test
(5k + 5)!
k +1
10 (3k + 3)!(2k + 2)!
= lim
k
(5k)!
k
10 (3k)!(2k)!
1
(5k + 5)(5k + 4)(5k + 3)(5k + 2)(5k + 1)
= lim
10 k (3k + 3)(3k + 2)(3k + 1)(2k + 2)(2k + 1)
=
1 55
>1
10 332 2
and so series diverges.
(b) By the ratio test,
9 k +1 (3k + 3)!(k + 1)!
(4k + 4)!
= lim
k
9 k (3k)!k!
(4k)!
(3k + 3)(3k + 2)(3k + 1)(k + 1)
= 9 lim
k (4k + 4)(4k + 3)(4k + 2)(4k + 1)
=9
33
<1
44
so series converges.
(c) By the ratio test,
4 k (k!)4
(c) (4k)!
1
50
K. A. Tsokos: Series and D.E.
4 k +1 ((k + 1)!)4
(4k + 4)!
= lim
k
4 k (k!)4
(4k)!
(k + 1)4
= 4 lim
k (4k + 4)(4k + 3)(4k + 2)(4k + 1)
1
= 3 <1
4
so series converges.
2
sin 2n
14. (a)
1
(b)
(a) For large n the general term behaves as
series
1
n
2
(c)
1
Solution
1 cos n cos
2
1
(n 1)
n
so we compare with the convergent
2n 2
. By the limit comparison test and using L’ Hôpital’s rule,
1
2
cos 2 2
3
2n = lim 2n
2n = ,
n
1
2
2
n2
n3
sin
lim
n
hence the series converges.
(b) We will be able to see when we study MacLaurin series that the general term will
1
behave as
so we compare with the convergent series 2 . By the limit
2
2n
1 n
comparison test and using L’ Hôpital’s rule,
1 cos
2 sin
sin
cos
2
2
n = lim n
n = lim
n = lim n
n = ,
lim
n
n
1
2
1
2 n 1
2 n
2
2
3
2
n
n
n
n
hence series converges.
(c) lim cos 2
n
(n 1)
= cos 2 = 1 0 and so series diverges by the divergence test.
n
15. The series of positive terms
un converges. Deduce that the series (a)
1
2
u
2
n
and
1
1
1
(b) un where > also converge. (c) Hence or otherwise prove that
n
2
1
u
1
the series n ( > ) converges.
2
1 n
Solution
51
K. A. Tsokos: Series and D.E.
(a) Since
u
n
converges lim un = 0 and so for sufficiently large n, un < 1 . Then
n
1
un2 < un and so
u
2
n
converges by the direct comparison test.
1
(b) Since (un and
1
n
2
1 2
1
2u
1
2u
) = un2 + 2 n < un2 + 2 (since n > 0 ) and
n
n
n
n
n
u
2
n
converges
1
also converges being a p-series with p = 2 > 1 we have that
1
2
1
un n converges as well by the comparison test.
1
2
1
1
u
u
=
n n (un2 + n 2 2 nn ) . All the series converge and thus so must
1
1
u
1
nn provided > 2 .
1
1
16. For what value(s) of p does the series p
converge?
2 n ln n
(c)
Solution
1
1
1
For p > 1 , p
< p and the series converges by comparison with p . For p = 1
n ln n n
2 n
1
that diverges as shown earlier (Set 3.3 Ex. 2(a)) by the integral
the series is 2 n ln n
1
1
test. For p < 1 , n p < n and so p
that diverges. Hence series converges for
>
n ln n n ln n
p > 1 and diverges for p 1 .
17. Examine the convergence or divergence of the two series (a)
1
(ln n)
p
and (b)
2
1
(ln n)
n
.
2
Solution
(a) The series clearly diverges for p 0 so we may assume that p is positive. In the text
we stated that for n sufficiently large, ln n < n . This can be proved by examining the
1/ 1
function f (x) = ln x x and seeing that it has a maximum at x = . Then,
1
1
1
1
> p . Choose such that p < 1 . Then series diverges by
> and so
p
n
(ln n)
ln n n
1
the direct comparison test since p diverges when p < 1 .
2 n
K. A. Tsokos: Series and D.E.
52
1
1
1
1
< n . Hence series
< and so
n
2
(ln n)
ln n 2
1
converges by the direct comparison test since n converges.
2 2
2
1 1
18. Using Euler’s result 2 =
find the sum .
2
6
1 n
1 (2n 1)
(b) Since ln n > 2 for n > e2 7.4 , then
Solution
1
n
2
=
1
1
1
1
1 1 2
1
1
= 2 + 2 =
+
=
+
2
2
2
2
4 1 n
6 n even n n odd n
1 (2n)
1 (2n 1)
1 (2n 1)
1
2 1 2 =
+
2
4 6
6
1 (2n 1)
1
2 1 2
=
(2n 1)2 6 4 6
1
=
2
8
1 1 1 1 1 1
+ + + + + + i.e. the sum of the
2 3 5 7 11 13
reciprocals of the prime numbers diverges. Use this fact to explain why the number
of prime numbers is infinite.
19. Euler has shown that the series
Solution
If the number of primes were finite the infinite series would converge. It diverges,
however, and so the number of primes is infinite.