Review ()
Winter 
Calculus II (-nyb-/,)
An old Test 
Another old Test 
Question . — Give the explicit solution of the initial value problem
et (1 − t)2
Question . — Express y as a function of x if
dy
x+1
=
,
dx xy + x
dx
= t(1 − x)2 ; x(0) = 0.
dt
and
y = −4 if x = 1.
Question . — One model for the spread of an epidemic is that the rate of spread is jointly
proportional to the number of infected people and the number of uninfected people. In an isolated
town of five thousand inhabitants, one hundred and sixty people have a disease at the beginning
of the week and twelve hundred have it at the end of the week. How long does it take for eighty
per cent of the population to become infected?
Question . — A tank contains 1000 L of pure water. Brine which contains 0.05 kg of salt per
litre of water enters the tank at a rate of 5 L/min, and brine which contains 0.04 kg of salt per litre
of water enters then tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains
from the tank at a rate of 15 L/min. How much salt is in the tank after t minutes? How much
salt is in the tank after one hour? When, if ever, does the concentration of salt in the tank exceed
0.04 kg/L? When, if ever, does the concentration of salt in the tank exceed 0.045 kg/L?
Question . — Evaluate each of the following limits, and simplify the results.
2
2
sin µ 1/µ
1
1
2 x
a. lim
− t
c. lim e−2x 1 +
b. lim
x→∞
t
µ
x
e
−
1
t→0
µ→0
Question . — Evaluate each of the following limits, and simplify the results.
√
x
n
o
1+x−e
2
a. lim x2 ln cos(1/x)
b. lim
c. lim (sin 5x)tan x
x→∞
x
x→0
x→ 1 π
Question . — Find the limit of each sequence or explain why it diverges.
(
)
n
o
n
o
(n!)2 sin(nn )
a. (−1)k k arccsc(k)
b. log(n2 + 1) − 2 log(3n − 2)
c.
(2n)!
1
for k > 0.
2 + ak
Question . — The sequence {ak } is defined by a0 = 1 and ak+1 = 1 −
a. Prove that 0 < ak+1 < ak 6 1 for all integers k > 0.
b. Explain why the sequence { ak } is convergent, and find its limit.
Question . — Find the sum of the series or explain why it is divergent.
∞ 2k−1
∞
∞ n
X
X
X
o
√
√
13j − 6
4
− 2 · 5k
a.
c.
b.
arctan( 2n + 1 ) − arctan( 2n + 5 )
j(j − 2)(j + 3)
15k
n=0
j=3
k=1
Question . — For each series, determine whether it is convergent or divergent. To earn full
credit, write√complete and proper solutions.
∞
∞
∞
X
X
X
csc2 (n)
5k 3 − 2k + 7
n3n
a.
b.
c.
√
3
(3n)!
ln n
n=1
n=2
k=0 k + 3k 5 − k + 7
Question . — For each series, determine whether it is absolutely convergent, conditionally
convergent or divergent. To earn full credit, display complete and proper solutions.
∞
∞
∞
∞
X
X
X
X
2 − 3j j/3
sin(k) (ln(k))3
(−1)n n3
1 · 3 · 5 · · · (2k − 1)
a.
b.
c.
d.
(−1)k
√
3
2j
+
5
2 · 4 · 6 · · · (2k)
k2 − k + 1
e n
k=1
n=0
j=1
k=1
Question . — Find the radius and interval of convergence of the power series
∞
X
tan(1/k)
k=2
2k (ln k)3/2
k=1
b. Show that if ak > 0 for k > 1, then
k=1
Question . — Find the sum of each series or else explain why it is divergent.
∞
2
X
3
8k − 2k − 15
1
9
1
27
1
81
1
a.
ln
b. +
−
+
+
+
−
+
+ ···
2
8 1 · 4 32 2 · 5 128 3 · 6 512 4 · 7
8k − 2k − 3
k=2
Question . — For each series, determine whether it is convergent or divergent. State which test
you are using and verify that the conditions for using it are satisfied.
!k 2
∞
∞
∞ n−2
X
X
X
π + e sin(m)
k−3
2
+ 5n+1
a.
22k
b.
c.
√
n + 7n−1
2
k
3
m +1
m=0
n=0
k=1
Question . — Determine whether each series is absolutely convergent, conditionally convergent
or divergent. Justify all assertions carefully.
∞
∞
∞
X
X
X
cos(πn) arccot(n)
sin(1/k)
(−3)3k (k!)3
b.
(−1)k
c.
a.
√
3
ln k
(3k)!
n
n=1
k=2
k=1
Question . — Find the radius and interval of convergence of the power series
∞
X
(−1)k ln k
k=1
2k k 1/5
1
1
< an < ,
2n
n
∞
√
P
Question . — Consider sequences {ak }, {bk } such that lim bk = 2, and
ak is convergent.
∞
P
Question . — a. Let an = ln(n + 1) − ln(2n + 1), and let {bn }n>1 satisfy b1 + · · · + bn = an , for n > 1.
∞
∞
P
P
Determine the value or limit, or explain the divergence, of {an }, {bn },
an and
bn .
n=1
n=1
( k)
n
o
k
b. Find the limit or explain the divergence of: i.
; ii. (−1)n csc(π−n ) 12 π − arcsec(πn+1 ) .
k!
(3x − 2)k .
Question . — Let {an } be a sequence of positive real numbers such that
(3x − 5)k .
a. Give an example of a pair of such sequences for which
2
∞
P
k=1
ak bk is convergent.
ak bk is divergent.
if
a. Show that if an+1 6 an for n > 1, then the series
n > 1.
∞
P
(∗)
(−1)n an is conditionally convergent.
n=1
b. Give an example of a sequence {an } which satisfies (∗), yet the series
∞
P
(−1)n an is divergent.
n=1
Question . — For 0 < r < 1, define %0 = r, %k+1 = %k + %k2 + %k3 + %k4 + · · · + %kn + · · · if this series
converges, and %k+1 = %k otherwise. Find the limit of the sequence {%k } (the result depends on r).
Review ()
Winter 
Calculus II (-nyb-/,)
Solution to Question . — a. If t = arccsc(k) and k > 1, then then k = (sin t)−1 , and so
Solutions to an old Test 
Solution to Question . — The differential equation is equivalent to
t→0+ sin t
dx
te−t
1
=
,
(1 − x)2 dt
(1 − t)2
and integrating gives
1
te−t
=
−
1−x 1−t
U −t
e (1 − t)
te−t
e−t
dt =
+ e−t + C =
+ C.
1−t
1−t
1−t
If x = 0 when t = 0, 1 = 1 + C, or C = 0. Therefore, 1 − x = et (1 − t), or x = 1 − et (1 − t).
Solution to Question . — Let y denote the number of people infected after t weeks. Then using
the given model there is a real number α such that
dy
= αy(1 − y/p),
where p = 5000
dt
is the population of the town. Separating variables and integrating gives
p
p/y 2 dy
p
= α,
i.e.,
log − 1 = C − αt,
− 1 = Ae−αt
or
p/y − 1 dt
y
y
(where A = eC , since 0 < y < p). Initially,
121
A = 5000
160 − 1 = 4 ,
38 .
so e−α = 363
Therefore, the eighty per cent of the inhabitants have become infected when
log 121
121 38 t = 10 − 1 = 1 ,
or after approximately
t=
weeks.
8
4 363
4
log(38/363)
and after one week
121 e−α = 5000 − 1 = 19 ,
1200
6
4
Solution to Question . — a. Combining terms and applying l’Hôpital’s Rule twice gives
1
et − t − 1
et − 1
et
1
lim
= lim
= lim t
= lim t
= 21 .
− t
e −1
t→0 t
t→0 t(et − 1)
t→0 e − 1 + tet
t→0 2e + tet
b. Notice that
sin µ
lim 1 +
−1
µ
µ→0
and
by two applications of l’Hôpital’s Rule and the fact that
lim
µ→0
sin µ
µ
1/µ2
sin µ
µ → 1 as µ → 0. Therefore,
Therefore,
= lim
and hence
1
t→0+ 2(1 + t)
2
2 x
lim e−2x 1 +
= exp −4 · 12 = e−2 .
x→∞
x
= 12 .
1
1
1 < 1−
< 1−
6 32 .
2
2 + ak+1
2 + ak
In other words, 12 < ak+2 < ak+1 6 23 , and so 0 < ak+2 < ak+1 6 1. By the Principle of Mathematical
Induction, 0 < ak+1 < ak 6 1 for every integer k > 0.
b. From the solution to Part a, it follows that the sequence {ak } is bounded and decreasing, so
it converges to some real number α such that 0 6 α < 1. By elementary limit laws,
1
1
α = lim ak+1 = lim 1 −
= 1−
.
2 + ak
2+α
√
This last equation is equivalent to α 2 + α = 1, or (2α + 1)2 = 5, so α = − 12 + 21 5.
Solution to Question . — a. Since
is divergent,
and
∞ X
k
2 31
is convergent,
k=1
1
(the ratio of the first series is 16
15 , which is larger than 1, and the ratio of the second series is 3 ,
which is positive and smaller than 1), it follows that the series
∞ 2k−1
∞ X
k 4
− 2 · 5k X 1 16 k
=
− 2 13
15
4
15k
k=1
is divergent.
b. If
√
= 6 e−1 .
1
1 − 1+t
2t
1
1
1 6
<
< 1,
3
2 + ak 2 + ak+1 2
k=1
t − log(1 + t)
4 4
+ log(1 + t) = −4 ·
.
t t2
t2
Now one application of l’Hôpital’s Rule and revision gives
t − log(1 + t)
= lim
t2
t→0+
Solution to Question . — a. Since a0 = 1 and a1 = 1 − 13 = 32 , it is clear that 0 < a1 < a0 6 1.
Next, if k > 0 is an integer and 0 < ak+1 < ak 6 1, then 2 < 2 + ak+1 < 2 + ak 6 3, so
= e,
−2x + x2 log (1 + 2/x) = −
lim
so by definition the sequence in question converges to zero.
1 16 k
4 15
c. If t = 2/x, then the expression in the limit is the exponential function applied to
t→0+
c. If 0 < ε < 2 and n > − log2 (ε), then
(n!)2 sin(nn ) 2
3
n
1
·
·
···
< 2−n < ε,
<
(2n)!
2n + 1 2n + 2 2n + 3 2n
∞ X
sin µ
sin µ − µ
cos µ − 1
− sin µ
µ −1
lim
= lim
= lim
= lim
= − 16 ,
µ→0
µ→0
µ→0 3µ2
µ→0 6µ
µ2
µ3
= 1.
Since lim{k arccsc(k)} is a non-zero real number, the sequence { (−1)k k arccsc(k) } is divergent.
b. Combining the logarithms and revising gives
2
n
o
1 + n−2
n +1
=
lim
log
= log 19 .
lim log(n2 + 1) − 2 log(3n − 2) = lim log
(3n − 2)2
(3 − 2n−1 )2
1
sin µ
µ −1
t
lim{k arccsc(k)} = lim
and
k=1
√
√
bn = arctan( 2n + 1 ) − arctan( 2n + 5 ),
√
√
Bn = arctan( 2n + 1 ) + arctan( 2n + 3 ),
then bn = Bn − Bn+1 , for n > 0, so
b0 + b1 + · · · + bn−1 = (B0 − B1 ) + (B1 − B2 ) + · · · + (Bn−1 − Bn ) = B0 − Bn .
√
7 π, and lim B = π, the sum of the series is
Since B0 = arctan(1) + arctan( 3) = 12
n
∞
X
n=0
7 π − π = − 5 π.
bn = lim(B0 − Bn ) = 12
12
Review ()
c. The general term of the series is
b. If
13j − 6
1
1
1
2
3
1
1
aj =
= +
−
=3 −
+2
− ,
j(j − 2)(j + 3) j j − 2 j + 3
j j +3
j −2 j
by resolving into partial fractions and rearranging. Hence, if
1
1
1
1
1
+
+
Aj = 3 +
+2
,
j j +1 j +2
j −2 j −1
then aj = Aj − Aj+1 , for j > 3. Since lim Aj = 0, it follows that
∞
X
13j − 6
= lim(a3 + a4 + · · · + aj+2 ) = A3 − lim Aj+3
j(j − 2)(j + 3)
j=3
= 3 31 + 41 + 51 + 2 1 + 12
= 107
20 .
Solution to Question . — a. If
√
ak =
5k 3 − 2k + 7
k+
√
3
3k 5 − k + 7
,
and
bk =
1
,
k 1/6
then
3
(n + 1)3n+3 (3n)!
(n + 1)2
a
1 3n
lim n+1 = lim
· 3n = lim
· 1+
= 3e < 1,
an
(3n + 3)!
3(3n + 2)(3n + 1)
n
n
P
so the ratio test implies that the series an is convergent.
2
c. Since csc (n) > 1 and ln n < n for n > 3,
cn =
csc2 (n)
1
1
>
> ,
ln n
ln n n
√
3
e n
,
1
bn = 2 ,
n
and
if
k(2k − 1) 2k − 1
kak
=
=
> 1,
(k − 1)ak−1 2k(k − 1) 2k − 2
then
ka3k
and
bk =
1
k 3/2
,
then, since | sin(k)| 6 1 for all k > 1,
(ln k)3
k2
|a |
lim k 6 lim
· 2
= 0.
1/2
bk
k −k+1
k
P
P
Since bk is a convergent p-series (p = 32 > 1), the limit comparison test implies that ak is
absolutely convergent.
kak > (k − 1)ak−1 > · · · > 1 · a1 = 12 ,
=
k(2k − 1)3
=
8k 3 − 12k 2 + 6k − 1
= 1−
4k 2 − 6k + 1
< 1,
8k 3 − 8k 2
8k 3 − 8k 2
(k − 1)a3k−1 (k − 1)(2k)3
so ka3k < (k − 1)a3k−1 < · · · < 1 · a31 = 18 , or 0 < ak < 81 k −1/3 , and hence lim ak = 0. So the alternating
P
P
series test implies that (−1)k ak is convergent. Therefore, (−1)k ak is conditionally convergent.
Solution to Question . — Let
αk =
If x , 35 then,
n > 3.
Solution to Question . — a. If
so
P
and thus ak > 12 k −1 . So the comparison test implies that ak is diverges with the harmonic series.
On the other hand, if k > 1 then 0 < ak < ak−1 (since ak > 0 and ak /ak−1 < 1), and
P −1
P
n is a divergent p-series (p = 1 6 1), the comparison test implies that an is divergent.
sin(k)(ln(k))3
ak
,
k2 − k + 1
(−1)n n3
|a |
n5
lim n = lim √
= 0.
3
bn
e n
P
P
Since bn is a convergent p-series (p = 2 > 1), the limit comparison test implies that an is
absolutely convergent.
c. If
√j
2 − 3j j/3
3j − 2 3
cj =
= > 1,
,
then
lim |cj | = lim
2j + 5
2j + 5 2
P
so the series cj is divergent by the root test. (Alternatively, |cj | > 1 if j > 7, and the vanishing
condition applies.)
d. If k > 1 and
1 · 3 · 5 · · · (2k − 1)
ak
a
2k − 1
ak =
,
then
<1
but
lim k = 1,
=
2 · 4 · 6 · · · (2k)
ak−1
2k
ak−1
an =
so the ratio test is inconclusive. However, if k > 1 then
then ak > 0 and bk > 0 if k > 0, and
√
√
√
5
a
k 1/6 5k 3 − 2k + 7
5 − 2k −1 + 7k −3
lim k = lim
=
lim
=
,
√
√
√
3
3
3
bk
3
5
−2/3
−4
−5
k + 3k − k + 7
k
+ 3 − k + 7k
P
which is a positive real number. Since bk is a divergent p-series (p = 61 6 1), the limit comparison
P
test implies that ak is divergent.
b. If
n3n
an =
,
(3n)!
Since
Winter 
Calculus II (-nyb-/,)
tan(1/k)
2k (log k)3/2
(3x − 5)k .
)
(
tan 1/(k + 1) |3x − 5|k+1
2k (log k)3/2
αk+1 = lim
lim
·
3/2
αk
tan(1/k)|3x − 5|k
2k+1 log(k + 1)
tan 1/(k + 1)
1
= 21 |3x − 5| lim
· lim tan(1/k)
log(1+1/k) 3/2
1 + log k
sin 1/(k + 1)
1/k
1
1
= 2 |3x − 5| lim
·
· 1+
1/(k + 1)
sin(1/k)
k
= 12 |3x − 5|.
P
The Ratio Test implies that αk is absolutely convergent if 12 |3x − 5| < 1, i.e., x − 53 < 23 , or
1 < x < 73 , and divergent if x < 1 or x > 73 . If x is 1 or 73 , then let
ak = |αk | =
then
tan(1/k)
(log k)3/2
and
bk =
1
;
k(log k)3/2
tan(1/k)
sin(1/k)
a
lim k = lim
= lim
= 1.
bk
1/k
1/k
Review ()
Winter 
Calculus II (-nyb-/,)
P
P
Since bk is a convergent logarithmic p-series (p = 32 > 1), the series |ak | is convergent by the
P
P
Limit Comparison Test. So αk is (absolutely) convergent if x = 1 or if x = 73 . Therefore, αk has
h
i
interval of convergence 1, 73 and radius of convergence 23 .
P
Solution to Question . — a. Let ak = (−1)k−1 k −1/2 ; ak is a (conditionally) convergent alter√
1
nating p-series (p = 2 , which satisfies 0 < p 6 1). Next, let bk = 2 + ak ; then since lim ak = 0, it
√
follows that lim bk = 2. Observe that
√
√
ak bk = 2 ak + a2k = 2 ak + k −1 ,
P√
P
P
for k > 1, so that
2 ak is convergent and k −1 is divergent. Therefore, ak bk is divergent.
√
b. If lim bk = 2 then there is a positive integer K such that 1 < bk < 2, for k > K. Since ak > 0
P
for k > 1, it follows that 0 6 ak bk < 2ak , for k > K. Since ak is convergent, the Comparison Test
P
Implies that ak bk is convergent.
and hence
(y + 1)2 = 2(x + log x) + C,
where the initial condition yields (−3)2 = 2+C, or C = 7. Solving for y (notice that y +1 < 0 because
of the initial condition) then gives
p
y = −1 − 2(x + log x) + 7.
Solution to Question . — Let q be the quantity of salt in the tank (measured in kilograms) at
time t (measured in minutes). Measured in kilograms per minute, salt enters the tank at a rate of
5
4
13
100 · 5 + 100 · 10 = 20
and leaves the tank at a rate of
Therefore,
dq 13
3 q,
= 20 − 200
dt
or
q
3 q.
· 15 = 200
1000
d
3
130
q − 130
3 = − 200 q − 3 .
dt
−3t/200 , and A = 130 , since initially the
This is an exponential differential equation, so q − 130
3 = Ae
3
tank contains pure water. Therefore, after t minutes there are
−3t/200 )
q = 130
3 (1 − e
−9/10 ) kilograms of salt in the tank.
kilograms of salt in the tank. After one hour there are 130
3 (1 − e
The concentration of salt in the brine exceeds 0.04 kg/L when there are more than 40 kg of salt
in the tank, i.e., any time after the point at which q = 40 (since q is an increasing function of t).
1 , or t = 200 log 13. So the concentration of salt in the brine
Solving q = 40 for t gives e−3t/200 = 13
3
200
exceeds 0.04 kg/L after 3 log 13 minutes.
The concentration of salt in the brine exceeds 0.045 kg/L when there are more than 45 kg of
salt in the tank. This never occurs, since 45 > 130
3 .
Solution to Question . — a. If t = 1/x, then one application of l’Hôpital’s Rule gives
(
)
ln(cos t) 1
sin t
1
lim x2 ln cos(1/x) = lim
=
lim
·
= 21 ,
2
x→∞
t
cos t
t2
t→0+
t→0+
since (sin t)/t → 1 and cos t → 1, as t → 0+ .
= e lim
x→0
x − (1 + x) log(1 + x)
log(1 + x)
= −e lim
2x
x→0
x2
= − 12 e.
c. As x → 21 π, t = sin(5x) − 1 → 0 and hence (1 + t)1/t → e. Arithmetical properties of limits
and two applications of l’Hôpital’s Rule give
n
o
sin(5x) − 1
5 cos(5x)
25 sin(5x)
lim (sin(5x) − 1) tan2 (x) = lim
= lim
= − 25
= lim
2 ,
2
x→ 12 π
x→ 21 π − sin(2x) x→ 12 π 2 cos(2x)
x→ 12 π cos x
x→ 21 π
Solution to Question . — Separating variables and integrating gives
dy
1
= 1+ ,
dx
x
x→0
of l’Hôpital’s Rule, arithmetical limit laws, and the fact that x−1 log(1 + x) → 1 as x → 0, gives
√
(
x
)
log(1 + x)
1+x−e
1
lim
= lim (1 + x)1/x
−
2
x
x(1 + x)
x→0
x→0
x
√
2
and therefore lim (sin 5x)tan x = e−25 .
Solutions to another an old Test 
(y + 1)
b. As lim (1+x)1/x = e, the numerator and denominator each vanish as x → 0. Two applications
Solution to Question . — a. First observe that
∞
X
n=1
bn = lim (b1 + b2 + · · · + bn ) = lim an = lim ln
n→∞
n→∞
Next, the vanishing condition implies that
∞
P
n=1
n→∞
n+1
= ln 12 = − ln 2.
2n + 1
an is divergent and that limn→∞ bn = 0.
b. i. Since
(k + 1)k+1 k!
1 k
·
= lim 1 +
= e > 1,
lim
(k + 1)! k k
k
ii. If t = π−n then
n
o
lim π−n csc(π−n ) = lim
it follows that
t
t→0+ sin t
lim
kk
= ∞.
k!
= 1,
and if x = πn+1 , then one application of l’Hôpital’s Rule gives
1
π − arcsec(π
lim 2
π−n
n+1 )
1
=
π − arcsec(x) 1
x
1
1
lim 2
lim √
= .
=
π x→∞
π x→∞ x2 − 1 π
x−1
Hence, since taking limits preserves products,
n
o 1
lim csc(π−n ) 21 π − arcsec(πn ) = ,
π
n
o
1
n
−n
and therefore the the sequence (−1) csc(π ) 2 π − arcsec(πn ) is divergent.
Solution to Question . — a. Since 8k 2 −2k −15 = (2k −3)(4k +5) and 8k 2 −2k −3 = (2k +1)(4k −3),
it follows that the general term of the series in question is
!
!
(2k − 3)(4k + 5)
(2k − 3)(2k − 1)
= Ak − Ak+1 ,
where
Ak = ln
,
ak = ln
(2k + 1)(4k − 3)
(4k − 3)(4k + 1)
for k > 2. Hence, the sum of the series is
∞
X
k=2
1 − ln 1 = ln 4 .
ak = lim(a2 + a3 + · · · + ak+1 ) = lim(A2 − Ak+2 ) = ln 15
15
4
Review ()
Winter 
Calculus II (-nyb-/,)
b. The general term of the series is
n
n
1
1
1
bn = 83 − 34 +
= 38 − 34 + 31
−
,
(n + 1)(n + 4)
n+1 n+4
c. The general term of the given series is (−1)k ak , where
for
n > 0,
and the sum of n terms of the series is
3 1 − − 3 n + 11 − 1 1 + 1 + 1
b0 + b1 + · · · + bn−1 = 14
18 3 n n + 1 n + 2 .
4
3 + 11 = 52 .
Therefore, the sum of the series is 14
18
63
k−3
k
!k 2
,
then ak > 0 if k > 3 and
!k )
(
√
3
lim k ak = lim 22 1 −
= 4/e3 ,
k
P
which is smaller than one, so the series ak converges by the Root Test.
2
2
b. If m > 1 then m + 1 < 4m and sin(m) > −1, so
π + e sin(m)
π−e
1
bm = √
>√
> 12 (π − e) · > 0.
2
2
m
m +1
4m
P
So the Comparison Test implies that the series bm diverges with the harmonic series.
c. Let
n
2n−2 + 5n+1
cn = n
and
dn = 75 .
3 + 7n−1
P
Then dn is a convergent geometric series (|r| = 57 < 1), cn > 0 and dn > 0 if n > 0, and
1 2 n
n−2
cn
2
+ 5n+1 7n
4 5 +5
lim
= lim
·
= lim n
= 35,
3 +1
dn
3n + 7n−1 5n
which is positive real number. Therefore,
P
7
7
cn is convergent by the Limit Comparison Test.
Solution to Question . — a. The general term of the given series is cos(πn)an = (−1)n an , where
arccot(n)
.
an =
√
3
n
If bn = n−4/3 , then an > 0 and bn > 0 for n > 1, and one application of l’Hôpital’s Rule gives
arccot(n)
a
n2
lim n = lim
= lim 2
= 1,
−1
bn
n
n +1
P
which is a positive real number. As bn is a convergent p-series (p = 43 > 1), the Limit Comparison
P
P
Test implies that an is convergent. Therefore, the series (−1)n an is conditionally convergent.
b. If
sin(1/k)
sin(1/k)
1
a
sin t
ak =
and bk =
,
then
lim k = lim
= lim
= 1,
ln k
k ln k
bk
1/k
t→0+ t
P
P
where t = 1/k, and ak , bk > 0 if k > 2. Since bk is a divergent logarithmic p-series (p = 1), ak
P
k
diverges by the Limit Comparison Test. So (−1) ak is not absolutely convergent. Next,
0 < sin 1/(k + 1) < sin(1/k)
and
ln(k + 1) > ln k > 0,
if k > 2,
P
so 0 < ak+1 < ak . Since lim{sin(1/k)} = 0 and lim{ln k} = ∞, lim ak = 0, so (−1)k ak is convergent
P
by the Alternating Series Test. Therefore, (−1)k ak is conditionally convergent.
33k (k!)3
,
(3k)!
and
Solution to Question . — If x , 23 and
!1/5 )
(
ln(k + 1)
k
α
= 21 |3x − 2|,
lim k+1 = 21 |3x − 2| lim
·
αk
ln k
k+1
2k k 1/5
P
so the Ratio Test implies that the series αk is absolutely convergent if 12 |3x − 2| < 1, i.e., 0 < x < 43 ,
P
and divergent if x < 0 or x > 43 . If x = 0 then αk = k −1/5 ln k > k −1/5 , if k > 3, and k −1/5 is
P
a divergent p-series (p = 15 6 1), so αk is divergent by the Comparison Test. If x = 43 then
αk = (−1)k k −1/5 ln k, so αk > 0 if k > 1 and lim αk = 0. Also
o
d n −1/5
k
ln k = k −6/5 − 15 k −6/5 ln k = 51 k −6/5 (5 − ln k),
dk
αk =
Solution to Question . — a. If
ak = 22k
(3k + 3)(3k + 3)
33 (k + 1)3
ak+1
=
> 1,
=
ak
(3k + 3)(3k + 2)(3k + 1) (3k + 2)(3k + 1)
P
so ak+1 > ak > 0 if k > 0, and the Vanishing Condition implies that the series (−1)k ak is divergent.
ak =
(−1)k ln k
(3x − 2)k ,,
then
which is negative if k > e5 , where {k −1/5 ln k} is decreasing. So the Alternating Series
Test
implies
i
P
P
that αk converges if x = 43 . Hence, the series αk has interval of convergence 0, 43 and radius
of convergence 23 .
P
P
Solution to Question . — a. Since 0 < 1/(2n) < an for n > 1 and 1/n diverges, an diverges
by the Comparison Test. On the other hand, 1/n > an > an+1 > 1/(2n + 2) > 0, so {an } is positive
P
and decreasing, and lim an = 0. Hence, (−1)n an is convergent by the Alternating Series Test.
P
n
Therefore, (−1) an is conditionally convergent.
b. Define the sequence {an } by

3



if n is odd, and


 5n
an = 


4


if n is even and positive.

5n
If n > 1, then
1
3
4
1
<
6 an 6
<
2n 5n
5n n
If k > 1, then
a2k−1 =
and hence
2n
X
(−1)ν aν =
ν=1
P
and
k=1
4
2
=
,
5 · 2k 5k
k=1
k
1
1
k−2
>
=
>
.
5k(2k − 1) 5k(2k − 1) 5(2k − 1) 5k
1/k diverges to ∞, it follows that
lim
n→∞
which implies that
a2k =
) X
n (
n
n n
X
o X
2
3
k−2
a2k − a2k−1 =
−
=
;
5k 5(2k − 1)
5k(2k − 1)
k=1
also,
Since
3
5(2k − 1)
P
2n
X
(−1)ν aν = lim
ν=1
(−1)n an is divergent.
k→∞
n
X
k=1
k−2
= ∞,
5k(2k − 1)
Review ()
Calculus II (-nyb-/,)
Solution to Question . — Since 0 < r < 1, it follows that
r
%1 = r + r 2 + r 3 + · · · =
.
1−r
Next,
r
r
r 2
r 3
1
,
if
< 1,
or
2r < 1,
%2 =
+
+
+ ··· =
1−r
1−r
1−r
1 − 2r
1−r
and %2 = %1 otherwise. Likewise,
2 3
r
r
r
1
r
%3 =
+
+
+ ··· =
,
if
< 1,
or
3r < 1,
1 − 2r
1 − 2r
1 − 2r
1 − 3r
1 − 2r
and %3 = %2 otherwise. And so on. The smallest positive integer such that %k+1 = %k satisfies
1
kr < 1 6 (k + 1)r,
or
k < 6 k + 1,
r
in which case %k+j = %k for j > 1, and the limit of {%k } is
r
r
l m ,
=
1 − kr 1 − 1 − 1 r
r
where for a real number x, dxe denotes the smallest integer which is not less than x, the so-called
ceiling of x. To answer this question it was not necessary to know anything about this notation,
only to deduce the displayed inequality defining k.
Winter 