The document provides an overview of the integral test for determining convergence or divergence of infinite series. 1) It introduces the integral test theorem, which states that if f is continuous, positive, and decreasing, the series of terms f(n) converges if and only if the integral of f from 1 to infinity converges. 2) Examples are provided to illustrate applying the integral test to test convergence of specific series, such as 1/n^2 and ln(n)/n. 3) It defines p-series as having terms of the form 1/n^p and uses the integral test to prove such series converge if p>1 and diverge if p≤1.

1. Chapter 11: Sequences, Series, and Power Series
Section 11.3: The Integral Test
Alea Wittig
SUNY Albany

2. Outline
The Integral Test
p-Series
Estimating the Sum of a Series

3. ▶ So far we have covered the following types of series and
corresponding methods to determine convergence:
(a) Telescoping series - Try to find a simple formula for the nth
partial sum sn and use
lim
n→∞
sn =
∞
X
n=1
an
(b) Geometric series - Write in the form
P
arn−1
and determine
if |r| < 1.
(c) Divergent series - If it is easy to tell that lim
n→∞
an ̸= 0 then
the test for divergence tells you the series
P
an diverges.
(d) Sum, Difference, Constant Multiple of a known convergent
series - use the last theorem from 11.2.

4. ▶ In the next few sections we will develop more methods to
determine if a series is convergent or divergent without
explicitly determining its sum.
▶ The subject of this lecture is the Integral Test.

5. The Integral Test

6. Consider the series
1
1
+
1
22
+
1
32
+ . . . =
∞
X
n=1
1
n2
▶
P 1
n2 can be viewed geometrically as the sum of rectangles
under the graph of f (x) = 1
x2 ,
where the length of the base of each rectangle is 1 and the
height is the value of f (x) = 1
x2 at the right endpoint of the
interval.

7. ∞
X
n=2
1
n2
<
Z ∞
1
1
x2
dx
=⇒
∞
X
n=1
1
n2
= 1 +
∞
X
n=2
1
n2
< 1 +
Z ∞
1
1
x2
dx

8. Z ∞
1
1
x2
dx = lim
t→∞
Z t
1
1
x2
dx
= lim
t→∞
−
1
x

10. t
1
= lim
t→∞
1 −
1
t
= 1 =⇒
∞
X
n=1
1
n2
< 1 + 1 = 2
▶ Since 0 <
∞
X
n=1
1
n2
< 2, we know that the series converges to
come positive value less than 2.
▶ Note: this series actually converges to π2
6 . The proof is a bit
difficult.

11. ▶ Next, consider the series
∞
X
n=1
1
√
n
=
1
√
1
+
1
√
2
+ . . .

12. ▶ This time the series is equal to the sum of the areas of the
rectangles lying above the curve 1
√
x
for x ≥ 1 and the height
is the value of the function f (x) = 1
√
x
at the left endpoint of
the interval.
Z ∞
1
1
√
x
dx = lim
t→∞
Z t
1
1
√
x
dx
= lim
t→∞
2x1/2

14. t
1
= lim
t→∞
2
√
t − 2
= ∞

15. Since
Z ∞
1
1
√
x
dx <
∞
X
n=1
1
√
n
and
Z ∞
1
1
√
x
dx = ∞
we know that the series
∞
X
n=1
1
√
n
is divergent.

16. The Integral Test
Theorem: Suppose f is continuous, positive, and decreasing on
[1, ∞) and an = f (n).
∞
X
n=1
an is convergent if and only if
Z ∞
1
f (x)dx is convergent
In other words:
(i) If
Z ∞
1
f (x)dx is convergent, then
∞
X
n=1
an is convergent.
(ii) If
Z ∞
1
f (x)dx is divergent, then
∞
X
n=1
an is divergent.

17. The Integral Test - Remarks
▶ Note that we don’t need the series to start at n = 1 or integral
at x = 1 to use the integral test.
▶ For example, to test the series
∞
X
n=4
1
(n − 3)2
we use
Z ∞
4
1
(x − 3)2
dx
▶ It is also not necessary for f to be always decreasing, just
ultimately decreasing, that is, decreasing for x greater than
some number N.
If
∞
X
n=N
an convergent =⇒
∞
X
n=1
an is convergent
so long as an is defined as a real number for n < N.

18. Example 1
1/3
Test the series
∞
X
n=1
1
n2 + 1
for convergence or divergence.
▶ First we must check that the function f (x) satisfies the
assumptions of IT:
▶ continuous
▶ positive
▶ decreasing

19. Example 1
2/3
▶ f (x) =
1
1 + x2
is continuous on (−∞, ∞) as it is a rational
function with no infinite discontinuities since x2 + 1 ̸= 0 for
any real x.
▶ It is easy to see that f (x) is positive on (−∞, ∞) since
1
x2+1
> 0 ⇐⇒ x2 + 1 > 0 ⇐⇒ x2 > −1 which is true for all
values of x.
▶ To determine if f is decreasing, we take the derivative:
d
dx
1
1 + x2
= −
1
(1 + x2)2
· (2x)
= −
2x
(1 + x2)2
< 0 ⇐⇒ −2x < 0 ⇐⇒ x > 0
So f is decreasing on (0, ∞).

20. Example 1
3/3
▶ Now we apply the integral test: let’s find out whether
R ∞
1
1
1+x2 dx converges or diverges:
Z t
1
1
x2 + 1
dx = tan−1
t

23. t
1
= tan−1
t − tan−1
(1)
= tan−1
t −
π
4
lim
t→∞
Z t
1
1
x2 + 1
dx = lim
t→∞
tan−1
t −
π
4
=
π
2
−
π
4
=
π
2
=⇒
∞
X
n=1
1
n2 + 1
converges by the integral test

24. Example 2
1/4
Determine whether the series
∞
X
n=1
ln n
n
converges or diverges.
▶ First we must check that the function f (x) satisfies the
assumptions of IT:
▶ continuous
▶ positive
▶ decreasing

25. Example 2
2/4
▶ The function f (x) =
ln x
x
is continuous on (0, ∞) since both
ln x and 1
x are and continuous on (0, ∞).
▶ The function f (x) = ln x
x > 0 whenever
▶ ln x > 0 and x > 0
▶ ln x > 0 ⇐⇒ x > e0
, ie, x > 1.
▶ ln x < 0 and x < 0
▶ This is not possible since ln x < 0 ⇐⇒ 0 < x < 1
▶ We also don’t care about negative values of x anyways since
our series starts at n = 1.

26. Example 2
3/4
▶ To determine if f is decreasing we take the derivative:
d
dx
ln x
x
=
x 1
x − ln x · 1
x2
=
1 − ln x
x2
< 0 ⇐⇒
1 < ln x ⇐⇒ for x > e✓
so for x > e the function is decreasing.
▶ We have found that f satisfies the assumptions of IT for
x > e. Rounding up to the next integer, f satisfies the
assumptions on [3, ∞).
∞
X
n=1
ln n
n
=
ln 2
2
+
∞
X
n=3
ln n
n
▶ So we use the integral test on the interval [3, ∞)

27. Example 3
4/4
Z t
3
ln x
x
dx =
Z ln t
ln 3
udu


u = ln x du = 1
x dx
x1 = 3 → u1 = ln 3
x2 = t → u2 = ln t


=
u2
2

32. ln t
ln 3
=
(ln t)2
2
−
(ln 3)2
2
=⇒ lim
t→∞
Z t
3
ln x
x
dx = lim
t→∞
(ln t)2
2
−
(ln 3)2
2
= ∞
So by the integral test
∞
X
n=1
ln n
n
diverges

33. p-Series

34. p−Series
▶ We call a series of the form
∞
X
n=1
1
np
a p-series.
▶ We have seen at the beginning of the lecture that the p- series
with p = 2 converged, while the one with p = 1
2 diverged.
▶ Now we will use IT to find all values of p for which a p−series
converges.

35. p−Series
▶ The function f (x) = 1
xp is clearly positive and continuous on
[1, ∞) for any value of p.
▶ Is it decreasing (or ultimately decreasing) on [1, ∞)?
▶ If p < 0 then f (x) =
1
xp
= x|p|
which is actually increasing
on [1, ∞), so we can’t use the integral test.
But since lim
n→∞
an = lim
n→∞
1
np
= lim
n→∞
n|p|
= ∞ ̸= 0
∞
X
n=1
1
np
diverges for p < 0 by the test for divergence
▶ If p = 0 then f (x) = 1 and
P∞
n=1
1
np =
P∞
n=1 1 which also
diverges by the test for divergence.

36. p-Series
▶ If p > 0 then the function f (x) = 1
xp is decreasing on [1, ∞):
d
dx
1
xp
=
d
dx
x−p
= −px−p−1
= −
p
xp+1
< 0 for x > 0
so we can use the integral test.

37. p-Series p > 0
Z t
1
1
xp
dx =







ln |x|

40. t
1
if p = 1
x1−p
1 − p

45. t
1
if p ̸= 1
=



ln |t| if p = 1
t1−p
1 − p
−
1
1 − p
if p ̸= 1
▶ If p = 1 then lim
t→∞
Z t
1
1
xp
dx = lim
t→∞
ln t = ∞ so this shows
the harmonic series diverges.

46. p Series - p > 0, p ̸= 1
▶ And if p > 0, and p ̸= 1 then
lim
t→∞
Z t
1
1
xp
dx = lim
t→∞
t1−p
1 − p
−
1
1 − p
=
1
p − 1
+
1
1 − p
lim
t→∞
1
tp−1
=
(
∞ if p < 1
0 if p > 1

47. p−Series Test
The p-series
∞
X
n=1
1
np
is convergent if p > 1 and divergent if p ≤ 1
(a) The series
X
n=1
1
n3
is convergent bc it is a p − series with p > 1.
(b) The series
∞
X
n=1
1
n1/3
is divergent b/c it is a p − series with p ≤ 1.

48. Estimating the Sum of a Series

49. Estimating the Sum of a Series
▶ Suppose we have determined
P∞
n=1 an converges, but not its
exact value, s.
∞
X
i=1
ai =
n
X
i=1
ai +
∞
X
i=n+1
ai
s = sn + Rn
▶ The partial sums sn approximate s.
▶ The remainder,
Rn =
∞
X
i=n+1
ai = an+1 + an+2 + . . .
is the error made when sn is used as an approximation for s.

50. Z ∞
n+1
f (x)dx ≤ Rn =
∞
X
i=n+1
ai ≤
Z ∞
n
f (x)dx

51. Remainder Estimate for the Integral Test
Suppose f (k) = ak, where f is a continuous, positive, decreasing
function for x ≥ n and
P
an is convergent. If Rn = s − sn, then
Z ∞
n+1
f (x)dx ≤ Rn ≤
Z ∞
n
f (x)dx

52. Example 4
1/
(a) Approximate the sum of the series
P 1
n3 by using the sum of
the first 10 terms. Estimate the Error involved in this
approximation.
s10 = 1 +
1
8
+
1
27
+
1
64
+
1
125
+ . . . +
1
1000
≈ 1.197532
▶ First off, we know this p − series is convergent since p = 3.
▶ Let’s compute the integral on the left of Rn:
Z ∞
n+1
f (x)dx =
Z t
11
1
x3
dx
= −
1
2x2

55. t
11
= −
1
2t2
+
1
2 · 112
=
1
242
−
1
2t2

56. Example 4
2/
=⇒
Z ∞
11
1
x3
dx = lim
t→∞
1
242
−
1
2t2
=
1
242
=⇒
R10 ≥
1
242
▶ Now let’s compute the integral on the right:
Z ∞
n
f (x)dx =
Z t
10
1
x3
dx
= −
1
2x2

59. t
11
= −
1
2t2
+
1
2 · 102

60. Example 4
3/
Z t
10
1
x3
dx =
1
200
−
1
2t2
=⇒
Z ∞
10
1
x3
dx =
1
200
Thus
1
242
≤ R10 ≤
1
200

61. Example 4
4/
(b) How many terms are required to ensure that the sum is
accurate to within .0005?
▶ Need to find the value n such that
Rn ≤
5
104
Z ∞
n
1
x3
dx ≤
5
104
=⇒ Rn ≤
5
104
Z ∞
n
1
x3
dx =
1
2n2
≤
5
104
⇐⇒
1
n2
≤
10
104
103
≤ n2
10
√
10 ≤ n
31.6 ≤ n
So for n ≥ 32 we have Rn ≤ .0005.

62. Example 4
In general, to find the number of terms n to ensure sn is accurate
to within α: (where the series satisfies the conditions of the
Remainder Integral Test.)
Z ∞
n
f (x)dx ≤ α =⇒ Rn ≤ α

63. Example 5
1/2
Use n = 10 to estimate the sum of the series
P∞
n=1
1
n3 .
Rn = s − sn =⇒
Z ∞
n+1
f (x)dx ≤ s − sn ≤
Z ∞
n
f (x)dx =⇒
sn +
Z ∞
n+1
f (x)dx ≤ s ≤ sn +
Z ∞
n
f (x)dx
s10 ≈ 1.197532
s10 +
1
242
≤ s ≤ s10 +
1
200
=⇒
1.197532 +
1
242
≤ s ≤ 1.197532 +
1
200
1.201664 ≤ s ≤ 1.202532

64. Example 5
2/2
Approximating s by the midpoint of this interval
s ≈ 1.2021
The length of the interval is 1.202532 − 1.201664 = .000868 so the
error of the approximation 1.2021 is at maximum half of this, so
error is less than .000868/2 < .0005.