8.3. SERIES OF REAL NUMBERS
477
Therefore, if the improper integral
∞
1
f (x) dx diverges, so does the series
∞
k=1 a k .
1
2
3
4
5
6
7
8
9
10
0.00
0.25
0.50
0.75
1.00
k
a k
1
2
3
4
5
6
7
8
9
10
0.00
0.25
0.50
0.75
1.00
k
a k
Figure 8.4: Comparing an improper integral to a series
What’s more, if we look at the right hand Riemann sums of f on [1, n] as shown at
right in Figure 8.4, we see that
∞
1
f (x) dx >
∞
k=2
a k .
So if
∞
1
f (x) dx converges, then so does
∞
k=2 a k , which also means that the series
∞
k=1 a k converges. Our preceding discussion has demonstrated the truth of the Integral
Test.
The Integral Test. Let f be a real valued function and assume f is decreasing and
positive for all x larger than some number c. Let a k = f (k) for each positive integer
k.
1. If the improper integral
∞
c
f (x) dx converges, then the series
∞
k=1 a k converges.
2. If the improper integral
∞
c
f (x) dx diverges, then the series
∞
k=1 a k diverges.
Note that the Integral Test compares a given infinite series to a natural, corresponding
improper integral and basically says that the infinite series and corresponding improper
integral both have the same convergence status. In the next activity, we apply the Integral
Test to determine the convergence or divergence of a class of important series.