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THM – INTEGRAL TEST for convergence/divergence
Suppose that 𝑓(𝑥) is continuous, positive and decreasing function on the interval [1, ∞) and that 𝑓(𝑛) = 𝑎𝑛 then
∞
∞
∑ 𝑎𝑛 and improper integral ∫ 𝑓(𝑥) 𝑑𝑥 either both converge or both diverge.
𝑛=1
1
Proof of Integral Test
1.
Assumption: 𝑓(1) = 𝑎1 , 𝑓(2) = 𝑎2 , 𝑓(3) = 𝑎3 , … . , 𝑓(𝑛) = 𝑎𝑛
Area under f(x) on interval [1, n]:
𝑛
𝐴 ≈ (1)𝑓(2) + (1)𝑓(3) + ⋯ + (1)𝑓(𝑛) = 𝑎2 + 𝑎3 + ⋯ + 𝑎𝑛 = ∑ 𝑎𝑛
𝑛=2
𝑛
𝐴 = ∫ 𝑓(𝑥) 𝑑𝑥
1
𝑛
𝑛
𝑎)
f(x) is decreasing → ∑ 𝑎𝑛 ≤ ∫ 𝑓(𝑥) 𝑑𝑥
𝑛=2
𝑛
∑ 𝑎𝑛 ≤ 𝑎1 + ∫ 𝑓(𝑥) 𝑑𝑥
𝑛=1
∞
1
suppose that ∫ 𝑓(𝑥)𝑑𝑥 is convergent meaning it has finite value, then 𝑎1 + ∫ 𝑓(𝑥) 𝑑𝑥 ℎ𝑎𝑠 𝑓𝑖𝑛𝑖𝑡𝑒 𝑣𝑎𝑙𝑢𝑒 "𝑀"
1
𝑐)
add 𝑎1 to both sides →
1
∞
b)
𝑛
𝑛
1
∞
since f(x) is positive: ∫ 𝑓(𝑥)𝑑𝑥 ≤ ∫ 𝑓(𝑥)𝑑𝑥
1
1
𝑛
𝑎) & 𝑏) & 𝑐) → partial sum 𝑠𝑛 = ∑ 𝑎𝑛 ≤ 𝑀 for all n, so sequence {𝑠𝑛 } is bounded above
𝑛=1
Also, 𝑠𝑛+1 = 𝑠𝑛 + 𝑎𝑛+1 ≥ 𝑠𝑛 because 𝑎𝑛+1 = 𝑓(𝑛 + 1) ≥ 0. Thus, the sequence {𝑠𝑛 } is increasing bounded sequence,
therefore it is convergent by the monotonic sequence theorem. This means that ∑ 𝑎𝑛 is convergent.
∞
∞
EMPHASIZE:
∑ 𝑎𝑛 ≤ 𝑎1 + ∫ 𝑓(𝑥) 𝑑𝑥
𝑛=1
→
∑ 𝑎𝑛 is convergent if ∫ 𝑓(𝑥) 𝑑𝑥 is convergent
𝑛=1
1
2.
Q.E.D.
∞
∞
EQ 1
1
𝐴 ≈ (1)𝑓(1) + (1)𝑓(2) + ⋯ + (1)𝑓(𝑛 − 1) = 𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛−1
𝑛
𝐴 = ∫ 𝑓(𝑥) 𝑑𝑥
1
𝑛
𝑛−1
∫ 𝑓(𝑥) 𝑑𝑥 ≤ ∑ 𝑎𝑛
𝑛=1
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
∑ 𝑎𝑛 → ∞ , and so does ∑ 𝑎𝑛 → ∞ because 𝑎𝑛 ≥ 𝑎𝑛−1
𝑛=1
𝑛=1
This test does NOT give the value of a series. It will only give the convergence/divergence of the series.
Q.E.D .
2
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 − 𝑠𝑛 satisfies relationship:
∞
∞
from here we can find upper and lower bound for sum s
what provides more accurate approximation to the sum of
the series than the partial sum 𝑠𝑛 does.
∫ 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
∞
𝑛
s = ∑ 𝑎𝑛
𝑠𝑛 = ∑ 𝑎𝑛
𝑛=1
𝑠 = 𝑠𝑛 + 𝑎𝑛+1 + 𝑎𝑛+2 + 𝑎𝑛+3 + ⋯ = 𝑠𝑛 + 𝑅𝑛
𝑛=1
Goal is to estimate remainder, 𝑅𝑛 , which represent an error made when 𝑠𝑛 is used to approximate the total sum 𝑠
∞
𝑅𝑛 = 𝑎𝑛+1 + 𝑎𝑛+2 + 𝑎𝑛+3 ≤ ∫𝑛 𝑓(𝑥) 𝑑𝑥
∞
∞
EXAMPLE: Consider ∑ 𝑎𝑛 = ∑
𝑛=1
𝑛=1
∞
𝑅𝑛 = 𝑎𝑛+1 + 𝑎𝑛+2 + 𝑎𝑛+3 ≥ ∫𝑛+1 𝑓(𝑥) 𝑑𝑥
1
𝑛3
1. Determine whether the series converges or diverges.
∞
Check whether ∫ 𝑓(𝑥)𝑑𝑥 is convergent or divergent. If it is convergent so is series.
1
∞
𝑏
∞
1
1
1
1
1
1
∫ 3 𝑑𝑥 = lim ∫ 3 𝑑𝑥 = lim (− 2 + ) =
→ ∑ 3 is convergent .
𝑏→∞
𝑏→∞
𝑥
𝑥
2𝑏
2
2
𝑛
1
𝑛=1
1
2. Estimate the sum by using upper bound. or : find the upper bound.
∞
∞
∞
1
1
1
1
3
∑ 3 = 1 + ∑ 3 ≤ 1 + ∫ 3 𝑑𝑥 = 1 +
=
𝑛
𝑛
𝑥
2
2
𝑛=1
𝑛=2
∞
→
∑
𝑛=1
1
1
3
≤
𝑛3
2
3. Approximate the sum of the infinite series by the partial sum, 𝑠𝑛 (n=5) and determine accuracy (estimate error 𝑅𝑛 ).
∞
if n is given:
𝑠 ≈ 𝑠𝑛 and size of the error: 𝑅𝑛 ≤ ∫ 𝑓(𝑥)𝑑𝑥
𝑛
𝑠 ≈ 𝑠𝑛
1
1
1
1
= 1 + +
+
+
= 1.18566
8
27
64
125
𝑏
error: 𝑅𝑛 ≤ lim ∫
𝑏→∞
5
1
1
𝑑𝑥 =
= 0.02
3
𝑥
50
Q.E.D
3
4. Accuracy, A, is given. Find how many terms in the sum are required to ensure that accuracy is A.
∞
𝑅𝑛 ≤ A, so you have to find n that will give that accuracy :
∫ 𝑓(𝑥)𝑑𝑥 ≤ 𝐴
𝑛
How many terms are required to ensure that the approximate sum is accurate to within 0.001?
𝑏
𝑅𝑛 ≤ lim ∫
𝑏→∞
𝑛
1
1
1
𝑑𝑥 =
→
≤ 0.001
𝑥3
2𝑛2
2𝑛2
→ 𝑛2 ≥ 500 → 𝑛 ≥ 22.4 → 23 terms
∞
5. Estimate the sum ∑ 𝑎𝑛 and find accuracy.
𝑛=1
∞
∞
𝑠𝑛 + ∫ 𝑓(𝑥)𝑑𝑥 ≤ 𝑠 ≤ 𝑠𝑛 + ∫ 𝑓(𝑥)𝑑𝑥
𝑛+1
𝑛
approximations:
𝑠 =
𝑅𝑖𝑔ℎ𝑡 𝑠𝑖𝑑𝑒 + 𝐿𝑒𝑓𝑡 𝑠𝑖𝑑𝑒
2
error is at the most half of the length of the interval =
∞
Estimate ∑
𝑛=1
𝑠5 = 1 +
1
𝑛3
𝑅𝑖𝑔ℎ𝑡 𝑠𝑖𝑑𝑒 − 𝐿𝑒𝑓𝑡 𝑠𝑖𝑑𝑒
2
let 𝑛 = 5
1
1
1
1
+
+
+
= 1.18566
8
27
64
125
1.19954 < 𝑠 < 1.20566
∞
∫
5
1
1
𝑑𝑥 =
= 0.02
3
𝑥
50
∞
∫
6
1
1
𝑑𝑥 =
= 0.01389
3
𝑥
72
estimate: s ≈ 1.2026
error: 0.00306
Less error then in problem 4. with 23 terms approximation
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.
EXCERSISE: WOW
http://college.cengage.com/mathematics/blackboard/content/larson/calc8e/calc8e_solution_
main.html?CH=00&SECT=a&TYPE=se
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