1
Suppose that 𝑓(𝑥) is continuous, positive and decreasing function on the interval [𝑘, ∞) and that 𝑓(𝑛) = 𝑎𝑛
then
∞
∞
1. If ∫ 𝑓(𝑥) 𝑑𝑥 is convergent so is ∑ 𝑎𝑛
𝑘
𝑛=𝑘
∞
∞
2. If ∫ 𝑓(𝑥) 𝑑𝑥 is divergent so is ∑ 𝑎𝑛
𝑘
𝑛=𝑘
Proof
For the sake of the proof we’ll do it with the series that starts at 𝒏 = 𝟏. Generalize it later for any 𝒌.
Let’s define continuous, positive and decreasing function 𝑓(𝑥) such that 𝑓(𝑛) = 𝑎𝑛
Assumption: 𝑓(1) = 𝑎1 , 𝑓(2) = 𝑎2 , 𝑓(3) = 𝑎3 , … . , 𝑓(𝑛) = 𝑎𝑛
Area under f(x) on interval [1, n]:
𝑛
𝐴 = ∫ 𝑓(𝑥) 𝑑𝑥
1
1. Underestimation of area
𝐴 ≈ (1)𝑓(2) + (1)𝑓(3) + ⋯ + (1)𝑓(𝑛)
= 𝑎2 + 𝑎3 + ⋯ + 𝑎𝑛
𝑛
= ∑ 𝑎𝑖
𝒊=𝟐
𝑛
𝑎)
f (x) is decreasing → ∑ 𝑎𝑖 ≤ ∫ 𝑓(𝑥) 𝑑𝑥 →
∞
b)
𝒊=𝟐
1
∞
𝑛
∑ 𝑎𝑖 ≤ 𝑎1 + ∫ 𝑓(𝑥) 𝑑𝑥
1
𝒊=𝟏
If ∫ 𝑓(𝑥) 𝑑𝑥 is convergent, then 𝑎1 + ∫ 𝑓(𝑥) 𝑑𝑥 has finite value "𝑀"
1
𝑐)
𝑛
𝑛
𝑛
∞
1
since f(x) is positive: ∫ 𝑓(𝑥) 𝑑𝑥 ≤ ∫ 𝑓(𝑥) 𝑑𝑥
1
1
𝑛
∞
𝑎) & 𝑏) & 𝑐) → partial sum 𝑠𝑛 = ∑ 𝑎𝑖 ≤ 𝑎1 + ∫ 𝑓(𝑥) 𝑑𝑥 = 𝑀
𝑖=1
for all 𝑛
1
⇒ sequence {𝑠𝑛 } is bounded above by 𝑴
Because the terms are positive 𝑠𝑛 ≤ 𝑠𝑛 + 𝑎𝑛+1 = 𝑠𝑛+1 ⇒ {𝑠𝑛 }∞
𝑛=1 is increasing bounded sequence.
2
By the monotonic sequence theorem sequence of partial sums {𝑠𝑛 }∞
𝑛=1 converges. This means that
series ∑𝑛𝑛=1 𝑎𝑛 is convergent.
𝐐. 𝐄. 𝐃.
2. Overestimation of area
𝐴 ≈ (1)𝑓(1) + (1)𝑓(2) + ⋯ + (1)𝑓(𝑛 − 1)
= 𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛−1
𝑛−1
= ∑ 𝑎𝑖
𝒊=𝟏
𝑛−1
𝑛
∫ 𝑓(𝑥) 𝑑𝑥 ≤ ∑ 𝑎𝑖
1
𝑖=1
∞
𝑛
If ∫ 𝑓(𝑥) 𝑑𝑥 is divergent, then ∫ 𝑓(𝑥) 𝑑𝑥 → ∞ as 𝑛 → ∞ because 𝑓(𝑥) ≥ 0
1
1
𝑛−1
so for 𝑛 → ∞
𝑛
∑ 𝑎𝑖 → ∞ , and so does ∑ 𝑎𝑖 → ∞ because 𝑎𝑛 ≥ 𝑎𝑛−1
𝑖=1
Q.E.D .
𝑖=1
This test does NOT give the value of a series.
It will only give the convergence/divergence of the series .
∞
● Consider the 𝐩 − 𝐬𝐞𝐫𝐢𝐞𝐬:
∑
𝑛=1
1
1
1
1
= 𝑝+ 𝑝+ 𝑝+⋯
𝑝
𝑛
1
2
3
𝑓(𝑥) =
1
𝑥𝑝
As 𝑓(𝑥)is continuous, positive, decreasing and EASILY INTEGRABILE, use integral test as test for convergence
1
1−𝑝
∞
1
∫ 𝑝 𝑑𝑥 =
1 𝑥
∞
{
𝑝>1
𝑝=1
The series converges for 𝒑 > 𝟏
The series diverges for 𝟎 < 𝒑 ≤ 𝟏
∞ 0<𝑝<1
∞
The divergent p − series ∑
𝑛=1
1
𝑛
with 𝑝 = 1 is called 𝐡𝐚𝐫𝐦𝐨𝐧𝐢𝐜 𝐬𝐞𝐫𝐢𝐞𝐬.
3
We saw that for convergent geometric series it is possible to find exact sum. For most of the series there
is a way to find weather the series is convergent or divergent, but it is impossible to find the sum of
converging series. We can estimate the value of series. We can just take a partial sum 𝒔𝒏 and use that
as an estimation of the value of the series 𝒔. There are now two questions that we should ask about this.
First, how good is the estimation? If we don’t have an idea of how good the estimation is then it really
doesn’t do all that much for us as an estimation.
Secondly, is there any way to make the estimate better?
So, let’s start with a general discussion about the determining how good the estimation is.
Let’s start with converging series:
𝑛
∞
∞
∑ 𝑎𝑖 = ∑ 𝑎𝑖 + ∑ 𝑎𝑖
𝑖=1
𝑖=1
⇒
𝑠 = 𝑠𝑛 + 𝑅𝑛
𝑅𝑛 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟
𝑖=𝑛+1
So, the remainder tells us the difference, or error, between the exact value of the series
𝒔 (which we don′ t know) and the value of the partial sum 𝒔𝒏 that we are using as the estimation of
the value of the series.
Of course we can’t get our hands on the actual value of the remainder because we don’t have the actual
value of the series. However, we can use get a pretty good estimate of the remainder provided we make
some assumptions about the series. Once we’ve got an estimate on the value of the remainder we’ll also
have an idea on just how good a job the partial sum does of estimating the actual value of the series.
THM – REMAINDER ESTIMATE FOR THE INTEGRAL TEST
Suppose 𝑓(𝑛) = 𝑎𝑛 , where 𝑓(𝑥) is continuous, positive and decreasing function on the interval
𝑥 ≥ 𝑛 and that ∑ 𝑎𝑛 is convergent. Remainder 𝑅𝑛 = 𝑠 − 𝑠𝑛 satisfies relationship:
∞
∫
∞
𝑓(𝑥) 𝑑𝑥 ≤ 𝑅𝑛 ≤ ∫ 𝑓(𝑥) 𝑑𝑥
𝑛+1
𝑛
∞
𝑠𝑛 + ∫
∞
𝑓(𝑥) 𝑑𝑥 ≤ 𝑠 ≤ 𝑠𝑛 + ∫ 𝑓(𝑥) 𝑑𝑥
𝑛+1
𝑛
Proof of Remainder Theorem
𝑠 = (𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 ) + 𝑎𝑛+1 + 𝑎𝑛+2 + 𝑎𝑛+3 + ⋯ = 𝑠𝑛 + 𝑅𝑛
∞
𝑅𝑛 = 𝑎𝑛+1 + 𝑎𝑛+2 + 𝑎𝑛+3 + ⋯ ≤ ∫ 𝑓(𝑥) 𝑑𝑥
𝑛
∞
𝑅𝑛 = 𝑎𝑛+1 + 𝑎𝑛+2 + 𝑎𝑛+3 + ⋯ ≥ ∫
𝑛+1
Q.E.D
𝑓(𝑥) 𝑑𝑥
4
Note:
1. The lower limit on the improper integral must be the same value that starts the series.
2. The function does not actually need to be decreasing and positive everywhere in the interval. All that’s really
required is that eventually the function (and hence series terms) is decreasing and positive. To see why this is true
let’s suppose that the series terms increase and or are negative in the range 1 ≤ 𝑛 ≤ 𝑁 and then decrease
and are positive for 𝑛 ≥ 𝑁 + 1. In this case the series can be written as
∞
𝑁
∑ 𝑎𝑛 = ∑ 𝑎𝑛 +
𝑛=1
𝑛=1
∞
∑ 𝑎𝑛
𝑛=𝑁+1
Now, the first series is nothing more than a finite sum (no matter how large N is) of finite terms and so will be finite.
So the original series will be convergent/divergent only if the second infinite series on the right is
convergent/divergent and the test can be done on the second series as it satisfies the conditions of the test.
It is important to note before leaving this section that in order to use the Integral Test the series terms MUST eventually be
decreasing and positive. If they are not then the test doesn’t work. Also remember that the test only determines the
convergence of a series and does NOT give the value of the series.