Section 11.3 The Integral Test
Ruipeng Shen
Jan 22
Brief Preview Let
∞
X
an be an infinite series with a general term an = f (n). Here f is
n=1
continuous, positive, decreasing function defined on [1, ∞). In order to determine whether the
series converges and estimate its sum, we may consider the improper integral of f (x).
Proposition 1. Assume n < N be positive integers. Let f (x) be a continuous, positive and
decreasing function defined on [n, ∞). Then the partial sum of the series with a general term
an = f (n) satisfies
Z N +1
Z N
N
X
ak <
f (x) dx <
f (x) dx.
n+1
k=n+1
n
The idea of proof The proof is illustrated by the following graphs. In each graph the area
of the blue region (union of rectangles) is equal to the partial sum. One can compare this area
with the integral by the geometric meaning of the integral.
y
y=f(x)
an+1
an+2
aN
n+1
n+2
n+3
N+1
Figure 1: comparison with
By considering the limit as N → ∞, we have
1
R N +1
n+1
f (x) dx
x
y
an+1
an+2
y=f(x)
aN
n
n+1
n+2
N
Figure 2: comparison with
RN
n
x
f (x) dx
Theorem 2 (The Integral Test and Estimate of Sums). Suppose that f (x) is a continuous,
∞
X
positive, decreasing function on [1, ∞) and let an = f (n). Then the series
an is convergent if
n=1
Z ∞
and only if the improper integral
f (x) dx is convergent. Furthermore, if they do converge,
1
then we have
Z
∞
f (x) dx ≤
sn +
n+1
∞
X
Z
∞
ak ≤ sn +
k=1
f (x) dx.
(1)
n
Remark 3. Let N be a positive integer. If f (x) is a continuous, positive, decreasing function
on [N, ∞) and an Z= f (n) for n ≥ N , then we can still use the integral test by considering the
∞
f (x) dx. In addition, the estimate (1) still holds for n ≥ N .
improper integral
N
Example 4. Determine whether the series
∞
X
1
converges or diverges.
2+1
n
n=1
Solution Because we know the improper integral
Z ∞
Z t
1
1
π π
π
dx
=
lim
dx = lim [arctan t − arctan 1] = − =
2
2
t→∞ 1 x + 1
t→∞
x +1
2
4
4
1
converges, we know the series is convergent.
Example 5. Determine for what values of p the p-series
Solution
Because we know the improper integral
Z ∞
1
dx
p
x
1
2
∞
X
1
is convergent.
p
n
n=1
converges if and only if p > 1 by a previous example, the p-series converges if and only if p > 1.
Example 6. Determine whether the series
∞
X
n=1
Solution
2n
converges or not.
+1
n2
Because the improper integral
Z t
Z ∞
2x
2x
dx = lim
dx = lim ln(t2 + 1) − ln 2 = ∞
2+1
2+1
t→∞
t→∞
x
x
1
1
diverges, we know the series is divergent.
∞
X
1
. How large is the error if we approximate the sum s by the partial
2
n
n=1
sum s100 ? How to obtain a better approximation with a little more effort?
Example 7. Let s =
Solution
By the estimate (1), we have
Z ∞
Z ∞
1
1
dx ≤ s − s100 ≤
dx.
2
2
101 x
100 x
By a basic calculation
Z
a
∞
1
dx = lim
t→∞
x2
Z
a
t
1
1 1
1
dx = lim − +
= ,
t→∞
x2
t
a
a
we have
1
1
≤ s − s100 ≤
.
101
100
Thus the error is approximately 1/100. A better approximation can be found by adding the
average of 1/101 and 1/100 to s100 . In fact the inequality above implies
201 1
1
201
1
≤
−
≤ s − s100 +
≤
⇐⇒ s − s100 +
.
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