Math 115 HW #3 Solutions - Colorado State University
Math 115 HW #3 Solutions
From ?12.2
20. Determine whether the geometric series
en 3n-1
n=1
is convergent or divergent. If it is convergent, find its sum. Answer: I can re-write the terms as
en
en-1
e n-1
3n-1 = e 3n-1 = e 3
.
Therefore, the series
en
e n-1
en
3n-1 = e
3
=e
, 3
n=1
n=1
n=0
where
the
second
equality
comes
from
shifting
the
index
by
one.
Since
e 3
<
1,
we
know
that
the geometric series
en
1
3
3
n=0
=
1-
e 3
=
. 3-e
Therefore, the given series converges and the sum is given by
en
en
3
3e
3n-1 = e
=e
=
.
3
3-e 3-e
n=1
n=0
24. Determine whether the series
k(k + 2) (k + 3)2
k=1
is convergent or divergent. If it is convergent, find its sum.
Answer: This series diverges. To see this, I will show that the terms in the sequence do not
go to zero:
k(k + 2)
k2 + 2k
lim
k
(k + 3)2
= lim
k
k2 + 6k + 9 .
Dividing numerator and denominator by k2 yields
lim
k
1 k2
1 k2
1
+
2 k
1
+
6 k
+
9 k2
=
lim
k
1
1 +
+
2 k
6 k
+
9 k2
= 1.
Therefore, using the nth term test (a.k.a. Test for Divergence), the series diverges.
1
38. Determine whether the series
n
ln
n+1
n=1
is convergent or divergent by expressing sn as a telescoping sum. If it is convergent, find its sum.
Answer: We can re-write the terms in the series as
n
ln
= ln(n) - ln(n + 1).
n+1
Therefore, the partial sum
sn = a1 + a2 + a3 + . . . + an = (ln(1) - ln(2)) + (ln(2) - ln(3)) + (ln(3) - ln(4)) . . . + (ln(n) - ln(n + 1)) = ln(1) + (- ln(2) + ln(2)) + (- ln(3) + ln(3)) + . . . + (- ln(n) + ln(n)) - ln(n + 1) = ln(1) - ln(n + 1) = - ln(n + 1).
Therefore,
n
ln
n+1
n=1
so the given series diverges.
=
lim
n
sn
=
lim (-
n
ln(n
+
1))
=
-,
48. Find the values of x for which the series
(x - 4)n
n=1
converges. Find the sum of the series for those values of x.
Answer: Notice that this is a geometric series, so the series converges when |x - 4| < 1, meaning that
3 < x < 5.
Moreover, for such values of x, the series converges to
1
1
=
.
1 - (x - 4) 4 - x
70. If an and bn are both divergent, is (an + bn) necessarily divergent? Answer: No. Let an = 1 for all n and let bn = -1. Then
an = 1 diverges
n=1
n=1
and
bn = (-1) diverges
n=1
n=1
2
However, certainly converges.
(an + bn) = (1 + (-1)) = 0 = 0
n=1
n=1
n=1
From ?12.3
16. Determine whether the series
n2 n3 + 1
n=1
is convergent or divergent.
Answer:
If
we
let
f (x)
=
x2 x3+1
,
then
the
terms
of
the
series
and
the
function
f
satisfy
the
hypotheses of the Integral Test, so the series will converge if and only if
x2
1
f (x)dx =
1
x3 + 1 dx
is finite. Letting u = x3 + 1, we have that du = 3x2dx, so I can re-write the above integral as
1 du 1
= ln |u| ,
3 u=2 u 3
2
which diverges since ln(u) as u . Therefore, the series diverges by the Integral Test.
22. Determine whether the series
1
n(ln n)2
n=2
is convergent or divergent.
Answer:
If
we
let
f (x)
=
1 x(ln x)2
,
then
the
terms
of
the
series
and
the
function
f
satisfy
the
hypotheses of the Integral Test, so the series will converge if and only if
1
2
f (x)dx =
2
x(ln x)2 dx
is finite.
Letting
u
=
ln x,
we
have
that
du
=
1 x
dx,
so
I
can
re-write
the
above
integral
as
du u=ln 2 u2 =
- u-1
1
=,
u=ln 2 ln 2
which is finite. Therefore, the series converges by the Integral Test.
3
30. Find the values of p for which the series
ln n np
n=1
is convergent.
Answer: When p 0 the terms in the series do not go to zero, so the series will diverge.
When
p
>
0,
the
function
f (x)
=
ln x xp
and
the
series
satisfy
the
hypotheses
of
the
Integral
Test, so the series will converge if and only if
ln x
f (x)dx =
1
1
xp dx
(*)
is finite.
When p = 1, I can write the above integral as
ln x dx.
1x
Letting
u
=
ln x,
we
have
du
=
1 x
dx,
so
this
is
equal
to
u2
udu =
,
u=0
20
which is infinite.
When p = 1, I will use integration by parts to evaluate the integral in (*). Letting u = ln x
and
dv
=
dx xp
,
I
have
that
u = ln x 1
du = dx x
dv = x-pdx
x1-p
v=
,
1-p
so the anti-derivative is given by
ln x
x1-p
1
1
xp
dx
=
(ln x) 1
-p
-
1-p
xp dx
x1-p
1 x1-p
= (ln x)
-
1-p 1-p1-p
=
x1-p
(1
- p) ln x (1 - p)2
-
1
.
Hence,
1
ln x xp dx
=
x1-p (1
- p) ln x - (1 - p)2
1
,
1
which is finite only when 1 - p < 0.
Therefore, the series converges when p > 1.
4
34. Find the sum of the series
n=1
1/n5
correct
to
three
decimal
places.
Answer: If we estimate the sum by the nth partial sum sn, then we know that the remainder
Rn is bounded by
1
1
n+1 x5 dx Rn n x5 dx.
This means that
1
11 1
Rn
n
x5 dx
=
- 4
x4
n
= 4n4 ,
so the estimate will be accurate to 3 decimal places when this expression is less than 0.001.
In other words, we want to know for what n is it true that
1
1
4n4
<
. 1000
Solving for n, we get that
n > 4 250 3.98.
So letting n = 4, we have that
1111
11
1
s4 = 15 + 25 + 35 + 45 = 1 + 32 + 243 + 1024 1.036
is an estimate of the sum of the series that is correct to three decimal places.
From ?12.4
18. Determine whether the series converges or diverges.
1 2n + 3
n=1
Answer: Use the Limit Comparison Test to compare this series to
1 n
.
We
see
that
lim
n
1 2n+3
1 n
n = lim
n 2n + 3
=
1 .
2
Therefore, since
1 n
diverges,
the
Limit
Comparison
Test
tells
us
that
the
series
1 2n+3
also
diverges.
26. Determine whether the series converges or diverges.
n+5
n=1 3 n7 + n2
Answer: Use the Limit Comparison Test to compare this series to
1 n4/3
.
We
see
that
lim
n
3 nn7++5n2
1 n4/3
(n + 5)n4/3 = lim
n 3 n7 + n2
n7/3 + 5n4/3
= lim
.
n 3 n7 + n2
5
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