Notes PDF
Introductory Problems
1. (a) recalling that ∫1x (1/t) dt = ln(x), use the ideas of the proof of the integral test to show ln(n+1) ≤ 1 + 1/2 + 1/3 + … + 1/n ≤ 1 + ln(n)
Solution
1. Taking n = 4 as an example, the sum of the areas of the rectangles in the figure below is 1+ 1/2 + 1/3 + 1/4. Certainly, this is larger than the area under the curve y = 1/x between x = 1 and x = n+1. Consequently,
ln(n+1) ≤ 1 + 1/2 + 1/3 + … + 1/n. (A)
On the other hand, recalling we’re taking n = 4, the next picture shows that 1/2 + 1/3 + … + 1/n ≤ ln(n) and consequently,
1 + 1/2 + 1/3 + … + 1/n ≤ 1 + ln(n) (B)
Combining inequalities (A) and (B) we see ln(n+1) ≤ 1 + 1/2 + 1/3 + … + 1/n ≤ 1 + ln(n)
1. (b) Show how slowly the harmonic series diverges by finding upper bounds for ∑n=1N1/n for N = 1,000, for N = 1,000,000, for N = 1,000,000,000, and for N = 1,000,000,000,000.
Answer
1. Upper bounds are: N: upper bound 1,000: 7.91 1,000,000: 14.82 1,000,000,000: 21.73 1,000,000,000,000: 28.64
Solution
1. Use the upper bound 1 + 1/2 + 1/3 + … + 1/N ≤ 1 + ln(N) to establish the upper bounds. N: 1 + ln(N) 1,000: 7.90776 1,000,000: 14.8155 1,000,000,000: 21.7233 1,000,000,000,000: 28.631
2. Suppose {an} is a decreasing sequence of positive terms with limn → ∞an= 0. Use ideas in the proof of the integral test to show that ∑n=1∞an converges if and only if ∑n=1(∞)2na2n converges.
Solution
1. Representing each term an as the area of a rectangle with base 1 and height an, we see:
That is, a1 + a2 + a3 + a4 + a5 + a6 + a7 + …= a1 + (a2 + a3) + (a4 + a5 + a6+ a7) + … ≤ a1 + 2a2 + 4a4 + … From this we see ∑an ≤ ∑2na2n (A) where the second series starts with n = 0. On the other hand, by looking at a similar picture we see that the series (starting from n = 0) ∑ 2na2n= a1 + a2 + a2 + a4 + a4 + a4 + a4 + a8 + a8 + a8 + a8 + … = (a1 + a2) + (a2 + a4) + (a4 + a4) + (a4 + a8) + (a8 + a8) + a8 + … ≤ 2a1 + 2a2 + 2a3 + 2a4 + 2a5 + … That is, ∑2na2n ≤ 2∑an (B) Thinking of the area interpretations of these series, from (A) we see that if ∑an diverges, then ∑2na2n diverges. From (B) we see that if ∑an converges, then ∑2na2n converges. From (A) we see that if ∑2na2n converges then ∑an converges. From (B) we see that if ∑2na2n diverges then ∑an diverges.
3.Does the series ∑n=1∞1/(n ln(n) ln(ln(n))) converge or diverge?
Answer
1. The series diverges.
Solution
1. First, for x ≥ 3, the function f(x) = 1/(x ln(x) ln(ln(x))) is decreasing to 0, so the terms an= f(n) satisfy the conditions of exercise 2.Then 2na2n= 2n/(2nln(2n)ln(ln(2n))) = 1/(n ln(2) (ln(n) + ln(ln(2)))) Now ln(ln(2)) < 0 so, 1/(n (ln(n) + ln(ln(2)))) > 1/(n ln(n)) From introductory exercise 4, we know ∑ 1/(n ln(n)) diverges, so ∑ 1/(n ln(2) ln(n)) diverges. Each term of the series ∑1/(n ln(2) (ln(n) + ln(ln(2)))) is larger than the corresponding term of the series ∑ 1/(n ln(2) ln(n)), so the former series must diverge. We elaborate on this type of argument in the Comparison Test section.