Calculus 2-The Integral and Comparison Tests, guaranteed 100% Pass

1


The Integral and Comparison Tests


The Integral Test:
1 1 1 1 1
Let’s examine the series: ∑∞
𝑛=1 = + + + +⋯
𝑛2 12 22 32 42

𝑓(𝑥) = 1/𝑥 2
1
12
1
1
22 1
32 1
42
52

0 1 2 3 4 5



If we exclude the first rectangle, notice:
∞ ∞ 𝑏
1 1
∑ 2 ≤ ∫ 2 𝑑𝑥 = lim ∫ 𝑥 −2 𝑑𝑥 = lim − (𝑏 −1 − 1−1 )
𝑛 1 𝑥 𝑏→∞ 1 𝑏→∞
𝑛=2

1
= lim − ( − 1) = 1.
𝑏→∞ 𝑏

Since the first term of the original series is 1 we have:
∞
1
∑ ≤1+1=2
𝑛2
𝑛=1
1
So ∑∞
𝑛=1 converges because the partial sums {𝑆𝑛 } are bounded and
𝑛2
increasing.

, 2


1
Now let’s look at ∑∞
𝑛=1 .
𝑛
1
= 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒
1
𝑦 = 1/𝑥
1

1
2
1
1/2 3 1
4




1 2 3 4 5


Notice that:
∞1 1
∫1 𝑑𝑥 ≤ ∑∞
𝑛=1 𝑛
𝑥


But we have:


∞ 𝑏
1 1
∫ 𝑑𝑥 = lim ∫ 𝑑𝑥 = lim (ln(𝑏) − ln(1)) = ∞
1 𝑥 𝑏→∞ 1 𝑥 𝑏→∞



1
So ∑∞
𝑛=1 diverges.
𝑛