Math 1B
§ 11.3 The Integral Test
In general, it is difficult to find the exact sum of a series. We were able to do this for geometric series
and telescoping series because in each case we could find a simple formula for the nth partial sums, sn.
But it usually isn’t easy to compute lim sn .
n →∞
In the next few sections we develop several tests that enable us to determine whether a series is
convergent or divergent without explicitly finding its sum. The first test involves improper integrals.
€
Monotonic Sequence Theorem:
Every bounded, monotonic sequence is convergent.
(A sequence that is increasing and bounded above is convergent. A sequence that is decreasing and
bounded below is convergent.)
∞
For an infinite series
∑a
n
of non-negative terms, the sequence of partial sums
n=1
s1 = a1
s2 = a1 + a2
s3 =€a1 + a2 + a3

is increasing.
∞
€ So, a series
∑a
n
of non-negative terms for which the partial sums sn = a1 + a2 + ...+ an
n=1
are all bounded from above by some constant M ( sn ≤ M ) must converge.
€
Consider
€
∞
1
∑n
2
.
€
n=1
1
Let’s estimate the area under the curve f (x) = 2 on the interval [1,∞) using the right endpoint
x
€
approximation:
€
Stewart – 7e
1
The area of the rectangles will be an underestimation of the actual area and the estimation is not quite
the same as sn . Notice however that the only difference is that we’re missing the first term. This
means we can do the following:
€
All the terms of the series are positive so the sequence of partial sums is increasing (monotonic). Also,
sn < 2 so the sequence of partial sums is bounded.
∞
∴ The sequence of partial sums is a convergent sequence ⇒
€
2
is convergent.
n=1
∞
€
1
∑n
Note: It can be shown that
1
∑n
n=1
2
=
π2
6
€
€
We have related a series to an improper integral that we could compute and it turns out that the
improper integral and the series have the same convergence.
€
∞
Consider
1
∑n.
n=1
1
Let’s estimate the area under the curve f (x) = on the interval [1,∞) using the left endpoint
x
€
approximation:
€
The area of the rectangles will be an overestimation of the actual area and the estimation is the same as
sn .
∞
€
∴ The sequence of partial sums is a divergent sequence ⇒
1
∑n
is divergent.
n=1
€
Stewart – 7e
2
€
€
Moral to the story:
∞
1
∑ n 2 converges because
n=1
∞
∫
1
1
dx converges.
x2
∞
1
∑ n diverges because
n=1
∞
1
∫ x dx diverges.
1
Which leads us to …
€
€
The Integral Test:
f ( n ) = an , then
€
€ function on [1,∞) and let
Suppose f is a continuous,
positive, decreasing
∞
∞
i) If
€
∫
f ( x ) dx is convergent, then
∑a
n
1
€
∫
1
€
∞
∞
ii) If
is convergent.
n=1
f ( x ) dx is divergent, then
€
∑a
n
is divergent.
n=1
Note:
€
€ by any positive integer.
1. The 1 can be replaced
2. f doesn’t need to be always decreasing; f must eventually be decreasing.
∞
3. This does not say that
∑a
n
n=1
2
∞
∞
and
1 π
For example, ∑ 2 =
but
6
n=1 n
€
∞
∫ f ( x )dx are the same (when they converge).
1
∞
∫
1
1
dx = 1.
x2
€
1
Example: Is the series ∑
convergent or divergent?
2
1+
9n
n=1
€
€
€
Stewart – 7e
3
∞
Example: Is the series
1
∑ n ln n
convergent or divergent?
n= 2
€
∞
Example: Is the series
∑ ne
−n 2
convergent or divergent?
n= 0
€
Stewart – 7e
4
We can use the Integral Test to get the following fact/test for some series.
∞
p-series: The p-series
1
∑n
p
is convergent if p > 1 and divergent if p ≤ 1.
n=1
Example: Determine whether each series is convergent or divergent.
€
€
∞
€
1
a) ∑ 7
n
n=1
€
∞
b)
∑
n=1
1
n
€
Stewart – 7e
5