More Challenging Problems: The integral test

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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.