9
Infinite Series
Copyright © Cengage Learning. All rights reserved.
9.3
The Integral Test and
p-Series
Copyright © Cengage Learning. All rights reserved.
Objectives
 Use the Integral Test to determine whether an
infinite series converges or diverges.
 Use properties of p-series and harmonic series.
Srīnivāsa Rāmānujan (22 December 1887 – 26 April 1920) was
an Indian mathematician and autodidact who, with almost no formal
training in pure mathematics, made extraordinary contributions to
mathematical analysis, number theory, infinite series and continued
fractions. Ramanujan's talent was said by the English mathematician
G.H. Hardy to be in the same league as legendary mathematicians
such as Gauss, Euler, Cauchy, Newton and Archimedes and he is
widely regarded as one of the towering geniuses in mathematics.
Born to a poor Brahmin family, Ramanujan first encountered
formal mathematics at age 10. He demonstrated a natural ability,
and was given books on advanced trigonometry written by S. L.
Loney. He mastered them by age 12, and even discovered theorems
of his own, including independently re-discovering Euler's identity.
He demonstrated unusual mathematical skills at school, winning
accolades and awards. By 17, Ramanujan conducted his own
mathematical research on Bernoulli numbers and the Euler–
Mascheroni constant. He received a scholarship to study at
Government College in Kumbakonam, but lost it when he failed his
non-mathematical coursework.
Ramanujan joined another college to pursue independent
mathematical research, working as a clerk in the AccountantGeneral's office at the Madras Port Trust Office to support himself. In
1912–1913, he sent samples of his theorems to three academics at
the University of Cambridge.
Only Hardy recognized the brilliance
of his work, subsequently inviting Ramanujan
to visit and work with him at Cambridge. He
became a Fellow of the Royal Society and a
Fellow of Trinity College.
Ramanujan was a devout Hindu and
would not eat meat, nor would he eat food
prepared in any containers that had ever held
any meat. Also, his wife who had stayed
behind in India, told of times when she had to literally put food in
his mouth as he would forget to eat for days, entranced by his
“sums.” Without his wife’s care and with limited suitable food
available in England, Ramanujan died of malnutrition, and possibly
liver infection, in 1920 at the age of 32.
For the next few lessons, we will be
concerned with testing many different
types of series for convergence.
We will not be particularly concerned
with finding the sums of the series that
do happen to converge.
If that bothers you,
you may be an engineer.
Feel free to compute the infinite
sums of convergent series in
your copious spare time.
…just don’t forget to eat…
The integral Test is the first of many tests for convergence.
It is elegant and simple, but unfortunately has limited application.
It only works if the expression can be integrated when 𝑥 replaces 𝑛
in the term that defines the given series.
Example:
1
→
2
𝑛
1
𝑑𝑥
2
𝑥
Integral Test is appropriate.
Example:
1
𝑛!
1
𝑑𝑥
𝑥!
Integral Test is NOT appropriate.
→
f  x
𝑎1
𝑎2 𝑎3 𝑎4 𝑎5
. . .
𝑎𝑛
Here’s a graph of a series with 𝑎𝑛 → 0, which therefore MIGHT converge.
The blue curve represents the equivalent expression where 𝑥 replaces 𝑛.
The terms of the series can form an upper sum.
If the integral diverges…so does the “larger” series.
The same terms can also form a lower sum.
∞
∞
𝑎𝑛 >
𝑛=1
𝑓(𝑥)𝑑𝑥
1
𝑎1
𝑎2 𝑎3 𝑎4 𝑎5 𝑎6
. . .
𝑎𝑛
Here’s a graph of a series with 𝑎𝑛 → 0, which therefore MIGHT converge.
The blue curve represents the equivalent expression where 𝑥 replaces 𝑛.
The terms of the series can form an upper sum.
If the integral diverges…so does the “larger” series.
The same terms can also form a lower sum.
If the integral converges…so does the series.
∞
∞
𝑎𝑛 >
𝑓(𝑥)𝑑𝑥
1
𝑛=1
∞
∞
𝑎𝑛 <
𝑛=2
𝑓(𝑥)𝑑𝑥
1
Because the series can be drawn as either an upper or
lower sum, we can conclude that whatever the integral
does (converges or diverges), the series does too.
To use the integral test, all terms of the series must be
positive and decreasing over the long run.
You also need to know how to integrate the expression 
Integral Test
Given 𝑎𝑛 is a decreasing positive series and
𝑓 𝑥 a continuous function where 𝑓 𝑛 = 𝑎𝑛 , then
∞
∞
𝑎𝑛
𝑛=1
and
𝑓 𝑥 𝑑𝑥
1
either both converge or both diverge.
∞
Ex: #1 Apply the integral test:
𝑛=1
Terms positive?
∞
1
1
𝑑𝑥
𝑥
Decreasing?
1
𝑛
Can we integrate it?

 ln x 1  
Since the integral diverges, so does the series.
∞
Ex: #2 Apply the integral test:
𝑛=1
Terms positive?
∞
1
Decreasing?

1
1
𝑑𝑥 
2
𝑥
x 1
1
𝑛2
Can we integrate it?
 0   1  1
Since the improper integral converges, so does the series.
∞
Ex: #3 Decide if the series converges:
Terms are positive, but are they
decreasing?
∞
1
𝑥
𝑒
𝑑𝑥 = ?
𝑥2
𝑥2
𝑢=
𝑑𝑢 = 2𝑥𝑑𝑥
1
2
∞
1
1
𝑑𝑢
𝑢
𝑒
1 −𝑢
=− 𝑒
2
𝑛=1
1
=
2
∞
1
𝑛
𝑒𝑛
2
∞
𝑒 −𝑢 𝑑𝑢
1
−1
−1
1
=
−
=
2 ∙ ∞ 2𝑒 2𝑒
Since the integral converges, so does the series.
Ex: #4 Apply the integral test to the decreasing
positive series:
𝑛=1
∞
∞
1
∞
ln 𝑥
1 2
𝑑𝑥 = ? 𝑢 𝑑𝑢 = 𝑢
𝑥
2
1
∞
ln 𝑛
𝑛
=∞
1
𝑢 = 𝑙𝑛𝑥
1
𝑑𝑢 = 𝑑𝑥
𝑥
Since the integral diverges, so does the series.
The only thing that can make these problems difficult
is if the integral itself is difficult.
Ex: #5 You Try:
Apply the Integral Test to the series
Solution:
The function
is positive and continuous for
To determine whether f is decreasing, find the derivative.
Solution
So, f'(x) < 0 for x > 1 and it follows that f satisfies the conditions
for the Integral Test.
You can integrate to obtain
So, the series diverges.
p-Series and Harmonic Series
p-Series and Harmonic Series
A second type of series has a simple arithmetic test for
convergence or divergence. A series of the form
is a p-series, where p is a positive constant. For p = 1, the series
is the harmonic series.
p-Series and Harmonic Series
A general harmonic series is of the form
In music, strings of the same material, diameter, and
tension, whose lengths form a harmonic series,
produce harmonic tones.
p-Series
1
Any series of the form
where p is a real number.
𝑝
𝑛
p-series converge if and only if 𝑝 > 1
Proof
If 𝑝 = 1, then
1
𝑛𝑝
is the divergent harmonic series.
Otherwise, the related integral,
∞
∞ 1
1
−𝑝+1
𝑑𝑥
=
𝑥
1 𝑥𝑝
−𝑝+1
1
−𝑝+1
If 𝑝 > 1, then −𝑝 + 1 < 0 and 𝑥
→ 0 as 𝑥 → ∞
1
In this case, the integral converges to
and the series converges.
𝑝−1
0 and 𝑥 −𝑝+1
If 𝑝 < 1, then −𝑝 + 1 >
→ ∞ as 𝑥 → ∞
In this case, the integral diverges and therefore the series diverges.
Q.E.D.
Ex: #6– Convergent and Divergent p-Series
Discuss the convergence or divergence of (a) the harmonic
series and (b) the p-series with p = 2.
Solution:
a. From Theorem 9.11, it follows that the harmonic series
diverges.
b. From Theorem 9.11, it follows that the p-series
converges.
∞
Ex: #7 Determine whether the
Series converges or diverges:
𝑛=2
1
𝑛 ln 𝑛
This series is similar to the harmonic series. If its terms were larger than those
of the harmonic series, we would expect it to diverge. Because its terms are
smaller, we are not sure what to expect. The function is positive and
continuous for x>2. To determine whether f is decreasing, find its derivative.
Since f   x   
1  ln x
x  ln x 
2
∞
2
𝑑𝑥
𝑥 ln 𝑥
2
 0 for x  2, use the integral test.
∞
=
2
1/𝑥
ln 𝑥
𝑑𝑥
 lim ln  ln x   2  lim ln  ln b   ln  ln 2    
b
b 
b 
The series diverges.
∞
Ex: #16 Use the integral test:
𝑛=3
∞
3
𝑑𝑥
𝑥 ln 𝑥 ln(ln 𝑥)
𝑢 = ln(ln 𝑥)
𝑑𝑥
𝑑𝑢 =
𝑥(ln 𝑥)
1
𝑛 ln 𝑛 ln(ln 𝑛)
𝑝
∞
𝑝
𝑑𝑢
1
−𝑝+1
=
=
𝑢
𝑝
𝑢
−𝑝 + 1
ln(𝑙𝑛3)
∞
ln(ln3)
(we will need a special case if p = 1)
1
1
= lim
−
𝑢→∞ (1 − 𝑝)𝑢 𝑝−1
(1 − 𝑝) ln(𝑙𝑛3)
This limit will be zero if 𝑝 > 1, but ∞ if 𝑝 < 1
∞
𝑑𝑢
if 𝑝 = 1 then we get
= 𝑙𝑛 ∞ − 𝑙𝑛 𝑙𝑛 𝑙𝑛3
𝑙𝑛(𝑙𝑛3) 𝑢
The series converges if and only if 𝑝 > 1.
=∞
𝑝−1
Day 1: Pg. 620 1-41 odd
Day 2: Pg. 621 45-57 odd, 79-89 odd
Day 3: MMM pg. 196 (9.2), 199 (9.3)