Math 166
Study guide 3
Directions: This is practice for Exam 3. The exam will be Tuesday, November 3 in class. It will cover
material in textbook sections 8.7, 10.1-10.6. Expect the exam to contain roughly 7 problems. Roughly 6
problems will be in the style of Problems 1-6 below. One problem will be in the style of Problems 7 and 8.
1. Compute the following improper integrals or show that they diverge
(a)
∞
Z
1
dx
x4
(b)
4
Z
0
(c)
∞
Z
−∞
dx
√
x
dx
1 + x2
(d)
Z
1
∞
x3 + 1
dx
x4 + 4x3 + 4x2 + 5x + 3
2. Compute the following limits
(a)
ln(2n3 + 1)
n→∞
n
lim
(b)
lim
n→∞
1
1+
n
−n
(c)
r
lim
n→∞
n
3n
n
(d)
lim n 21/n − 1
n→∞
3. Find the sums of the following series
(a)
∞
X
1
(2n
−
3)(2n
− 1)
n=3
(b)
∞
X
−2
n(n
+ 1)
n=2
(c)
∞
X
e−n
n=2
(d)
∞ X
5
n 6
+ (−1) n
4n
7
n=1
(e)
∞
X
1
n=0
2n+2
4. Determine whether the following series converge.
(a)
∞
X
1
√
n
n=1
(b)
∞
X
ln n
n
n=1
(c)
∞
X
2n
3 + 4n
n=1
(d)
∞
X
n=1
(n2
n(n2 + 1)
+ 1)(n4 − 1)
(e)
∞
X
n(n + 1)
n2 − 1
n=1
(f)
∞
X
(ln n)3
n5
n=1
5. Determine whether the following series converge
(a)
∞
X
2n
n!
n=1
(b)
∞
X
n2 e−n
n=1
(c)
∞
X
n5n
(2n + 3) ln(n + 1)
n=1
(d)
∞
X
7
(2n
+
5)n
n=1
(e)
n
∞ X
4n + 3
n=1
3n − 5
(f)
−n
∞ X
1
1+
n
n=1
6. Which of the following series converge? Of those that converge, which converge absolutely?
(a)
∞
X
1
(−1)n √
n
n=1
(b)
∞
X
(−1)n
n=1
ln n
n
(c)
∞
X
(−1)n
n=1
n4
n2 + 1
+ n2 + 5
(d)
∞
X
(−1)n+1
n=1
3+n
5+n
Error estimates for series with positive terms
Let an = f (n) P
for all n ≥ 1, where f is a continuous,
Pnpositive, decreasing function defined on [1, ∞).
∞
Suppose
that
a
converges
to
S.
Let
s
=
k
n
k=
k=1 ak . We showed that the remainder Rn =
P∞
a
satisfies
the
inequality
k
k=n+1
Z
∞
Z
∞
f (x) dx ≤ Rn ≤
n+1
f (x) dx.
n
One should note that Rn = S − sn , which is why it makes sense to think of Rn as the error in using
sn as an estimate for S.
Consider the following example. Let f (x) = 1/x2 and consider
P∞
k=1
1/k 2 . One can easily compute
s10 = 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 + 1/49 + 1/64 + 1/81 + 1/100
P∞
with a calculator and think of s10 as an approximation for the entire sum S = k=1 1/k 2 . However,
this approximation is made much more meaningful if one can say something about the error R10 in
the approximation. The above inequality does this for us: it tells us that
Z
∞
R10 ≤
10
b
1
1
1
dx
−1 =
lim
−x
=
lim
−
−
.
=
b→∞
b→∞
x2
b 10
10
10
One can also turn this around and ask for n so that the error Rn is at most some small number like
10−5 . Note that
Z b
Z ∞
dx
1
1
1
dx
=
lim
=
lim
−
−
=
Rn ≤
2
2
b→∞ n x
b→∞
x
b n
n
n
So if we can find n so that 1/(n) ≤ 10−5 we will be done. Of course n ≥ 105 works.
7. Consider the convergent infinite series S =
P4
(a) Compute s4 = n=1 n13 .
P∞
1
n=1 n3 .
(b) Think of s4 as an approximation for S. Find an upper bound for the error in this approximation.
(c) Find the smallest integer n so that the error in using sn as an approximation for S is less than
10−7 .
Error estimates for alternating series
The previous method only works for error estimates in series with positive terms. For alternating
series,
P∞
we have the following error estimate. Consider a convergent alternating series of the form n=1 (−1)n an
converging to S. Again, let Rn = S − sn denote the error in using sn as an approximation for S. Then
|Rn | ≤ an+1 .
P∞
n+1 1
Consider the alternating series
n=1 (−1)
n2 . This series of course converges (by the absolute
convergence test), say to S. One can use a calculator to compute
s4 = 1 −
1
1 1
+ − .
4 9 16
We can think of s4 as an approximation for S. We’d like to know the error in this approximation.
This is precisely R4 = S − s4 . The issue is that we don’t know what S is. However, we know that
|R4 | ≤ a5 = 1/52 = 1/25. So the error in using s4 as an estimate for the entire sum S is at most 1/25.
8. Consider the convergent infinite series S =
P4
(a) Compute s4 = n=1 (−1)n n14 .
P∞
n 1
n=1 (−1) n4 .
(b) Think of s4 as an approximation for S. Find an upper bound for the error in this approximation.
(c) Find the smallest integer n so that the error in using sn as an approximation for S is less than
10−9 .