Page 1
Math 152-copyright Joe Kahlig, 15a
Section 10.3: The Integral and Comparison Tests; Estimating Sums
Note: In this section all series have positive terms.
This graph was used to discuss the Harmonic
series in section 10.2. We saw that
∞
X
1
n=1
n
>
Z∞
1
dx
x
1
and concluded that the Harmonic series diverged.
Here are the graph for the function f (x) =
∞ 1
P
1
. Does the series
converge or diverge?
2
2
x
n=1 n
Page 2
Math 152-copyright Joe Kahlig, 15a
The Integral Test Suppose f is continuous, positive, decreasing function on [1, ∞), or on
[A, ∞), and let an = f (n). Then the series
Z∞
∞
X
an is convergent if and only if the improper integral
n=1
f (x) dx is convergent. In other words:
1
(a) If
Z∞
f (x) dx is convergent, then
(b) If
an is convergent.
n=1
1
Z∞
∞
X
f (x) dx is divergent, then
∞
X
an is divergent.
n=1
1
Example: Assume that
∞
X
an is a series where f (n) = an and
n=1
Does the series
∞
X
an converge to L?
n=1
Example: Do these series converge or diverge?
A)
∞
X
n2
1 + n2
n=1
Z∞
1
f (x) dx converges to the number L.
Math 152-copyright Joe Kahlig, 15a
∞
X
B)
1
1 + n2
n=1
C)
∞
X
(ln(n))2
n=1
n
Page 3
Page 4
Math 152-copyright Joe Kahlig, 15a
The Comparison Test (Strict Comparison): Suppose that
positive terms.
(a) If
(b) If
X
X
bn is convergent and an ≤ bn for all n, then
bn is divergent and an ≥ bn for all n, then
Example: Do these series converge or diverge?
A)
B)
∞
X
n=1
5n3
∞
X
5n
n=1
6
+ n2 + 1
1
−2
X
X
X
an and
X
an is also convergent.
an is also divergent.
bn are series with
Page 5
Math 152-copyright Joe Kahlig, 15a
Limit Comparison Test(LCT): Suppose that
lim
n→∞
an
=L≥0
bn
X
an and
X
bn are series with positive terms and
If L > 0 then both series converge or both series diverge.
If L = 0 and
X
If L = ∞ and
X
bn converge, then
bn diverge, then
X
X
an converge.
an diverge.
(Note: This test is slightly different that the test given in the book.)
Example: Do these series converge or diverge?
A)
B)
∞
X
n=1
5n
∞
X
√
n=1
1
−2
n2
5
+ 2n − 7
Page 6
Math 152-copyright Joe Kahlig, 15a
C)
∞
X
3n2 + 5n
n=1
2n (n2 + 1)
X
Remainder Estimate For The Integral Test Suppose that
an converges by the Integral Test
to the number s. If sn is a partial sum that approximates this series, define Rn to be the remainder,
i.e. sn + Rn = s, then we get the following bounds on the remainder and the sum:
Z∞
n+1
f (x) dx ≤ Rn ≤
Z∞
n
f (x) dx
sn +
Z∞
n+1
f (x) dx ≤ s ≤ sn +
Z∞
n
f (x) dx
Math 152-copyright Joe Kahlig, 15a
Example: For
∞
X
30
i=1
i4
Page 7
.
A) Find the bounds on R10
B) What bounds are on the sum for this series?
C) If you wanted the error to be less than 0.005, what is the smallest value of n should you use?
Page 8
Math 152-copyright Joe Kahlig, 15a
Example: For
∞
X
3 + cos(i)
i=1
i4
.
A) Show the series converges.
B) Estimate the error on s6 .