May 12, 2006
Lecturer: Dr Martin Kurth
Trinity Term 2006
Problem Sheet 3 – Solutions
1. (a)
∞ µ ¶n−1
X
1
n=2
2
=
∞ µ ¶n−1
X
1
2
n=1
=
=
1
1−
1
−1
−1
1
2
(b)
¶n
∞ µ
X
1
−
4
n=1
= −
¶n−1
∞ µ
1
1X
−
4 n=1
4
1
1
¡ ¢
4 1 − − 41
1
= −
5
= −
(c)
∞
X
2n
n=1
µ ¶n−1
1
4
=
=
2
2
∞
X
2n−1
n=1
∞ µ
X
n=1
=
2
=
4
1
1−
1
2
µ ¶n−1
1
4
¶n−1
1
2
(d)
∞
X
n=1
n n−1
2 3
µ ¶n+1
1
8
=
=
=
=
1
µ ¶n−1
∞
2 X n−1 n−1 1
2
3
64 n=1
8
µ
¶
∞
n−1
1 X 3
32 n=1 4
1 1
32 1 −
1
8
3
4
2. (a) Direct comparison test.
and as
P∞
1
1
n3
cos2 (n)
1
≤ 3,
n3
n
2
P∞
converges, so does n=1 cosn3(n) .
(b) Direct comparison test again.
and as
1
n=1 n
P∞
1
2 − sin n
≥ ,
n
n
P∞
n
diverges, so does n=1 2−sin
.
n
(c) Yoohoo! Integral test! Define a decreasing function f by
f (x) =
e−2x
4
and calculate
∞
Z
¯b
1
lim e−2x ¯1
8 b→∞
1
,
8e2
f (x)dx = −
1
=
and as the integral converges, so does the series.
(d) Integral test again. First we follow the hint and calculate
¡ ¢
µ ¶
sinh x1
d
1
cosh
=−
.
dx
x
x2
Now we define our decreasing function f ,
¡ ¢
sinh x1
f (x) =
x2
and calculate
Z
∞
f (x)dx
= − lim cosh
b→∞
1
= cosh(1) − 1,
µ ¶¯b
1 ¯¯
x ¯1
and as the integral converges, so does the series.
3. First we have to find the points where the graphs of f and g intersect.
f (x) = g(x)
2
=⇒ x = x2
ie x = 0 or x = 1. Now we have to apply our washer cross section formula:
Z 1³
´
2
2
V = π
[f (x)] − [g(x)] dx
0
1
= π
Z
1
= π
Z
= π
µ
0
0
=
³
¡
−
7
π
15
£
¤2 ´
2
[x + 1] − x2 + 1
dx
¢
−x4 − x2 + 2x dx
¶¯1
¯
x5
x3
−
+ x2 ¯¯
5
3
0
4. As the series is convergent, we must have
lim an = 0.
n→∞
Now split the sequence {an } into its P
positive and negative subsequences
∞
{cn }, P
cn ≥ 0 and {dP
n }, dn < 0. As
n=1 is not absolutely convergent,
∞
∞
d
must
diverge.
Now we can construct our
c
and
both
n
n
n=1
P∞ n=1
rearranged series
b
in
the
following
way:
Take exactly as many
n
n=1
terms from {cn } as you need to make the sum just greater than L. Then
add terms from {dn }, such that the sum is (just) less than L. Continue
this procedure until infinity. After each step, the difference
¯
¯
n
¯X
¯
¯
¯
²n = ¯
bk − L¯
¯
¯
k=1
is less than or equal to the absolute value of the last term included in the
series. As limn→∞ an = 0, we have limn→∞ ²n = 0, and thus
∞
X
n=1
3
bn = L.