Category Archives: Numerical sequences and series Intro

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.

 

 

Numerical Sequences and Series: The ratio and root tests

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More Challenging Problems

Determine in the series a1 + a2 + a3 + … converges or diverges.

 

1. an = n2/3n

Answer

1. The series converges.

Solution

1. Use the Ratio test:                                                                                                                                   limn → ∞an+1/an                                                                                                                                                                                                = limn → ∞((n+1)2/3n+1)/ (n2/3n)                                                                                                                    = limn → ∞((n+1)2/n2)⋅(3n/3n+1)                                                                                                                      = limn → ∞((n+1)/n)2⋅(1/3)= 1/3                                                                                                                   So the series converges by the ratio test.

 

2. an= 3n/nn

Answer

1. The series converges.

Solution

1. The exponent n suggests the Root Test:                                                                                           limn → ∞(an)1/n                                                                                                                                                = limn → ∞(3n/nn)1/n                                                                                                                                           = limn → ∞ 3/n = 0                                                                                                                                           So the series converges by the Root Test.

 

3. an = n/(ln(n)n)

Answer

1. The series converges.

Solution

1. The exponent n in the denominator suggests the Root Test:                                                           limn → ∞(an)1/n                                                                                                                                             = limn → ∞(n/(ln(n)n))1/n                                                                                                                                    = limn → ∞ (n1/n)/ln(n)                                                                                                                                     For the n1/n factor, write y = x1/x, so ln(y) = ln(x)/x. Applying l’Hôspital’s rule, we see ln(y) → 0, so y → 1. Then with the ln(n) in the denominator, we see limn → ∞(an)1/n= 0, so the series converges by the Root Test.

 

4. an = (2n/(n + 1))n

Answer

1. The series diverges.

Solution

1. The exponent n suggests the Root Test:                                                                                           limn → ∞(an)1/n                                                                                                                                                   = limn → ∞((2n/(n + 1))n)1/n                                                                                                                             = limn → ∞ 2n/(n + 1) = 2                                                                                                                              So the series diverges by the Root Test.

 

5. an= n!/en

Answer

1. The series diverges.

Solution

1. he factorial suggests we try the Ratio Test:                                                                                       limn → ∞an+1/an                                                                                                                                                                                                = limn → ∞((n+1)!/en+1)/ (n!/en)                                                                                                                      = limn → ∞((n+1)!/n!)⋅(en/en+1)                                                                                                                        = limn → ∞ (n+1)/e So the ratio an+1/an → ∞ as n → ∞ and we see the series diverges by the Ratio Test.

 

6. an = n!/nn

Answer

1. The series converges.

Solution

1. Here we have both a factorial and an exponent n. In case you were thinking that factorial always means Ratio Test and that exponent n always means Root Test, this problem shows that rule cannot be applied to all series. Our intuition suggests we won’t have much luck with roots of factorials, but maybe we can handle ratios of exponents. So let’s try the Ratio Test:                 limn → ∞an+1/an                                                                                                                                                = limn → ∞((n+1)!/(n+1)n+1) / (n!/nn)                                                                                                               = limn → ∞((n+1)!/n!)⋅(nn/(n+1)n+1)                                                                                                                 = limn → ∞(n+1)⋅(nn/(n+1)n+1)                                                                                                                        = limn → ∞nn/(n+1)n                                                                                                                                        = limn → ∞(n/(n+1))n                                                                                                                              Recalling that: limn → ∞(1 + 1/n)n = we see: limn → ∞an+1/an= 1/e < 1 and so the series converges by the Ratio Test.

 

 

Numerical Sequences and Series: The alternating series test

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More Challenging Problems

In problems 1 – 4 determine if the series converges absolutely, converges conditionally, or diverges.

 

1. 1. 1 – 1/3 + 1/9 – 1/27 + 1/81 – …

Answer

1. The series converges absolutely.

Solution

1. To test absolute convergence, we test the series: 1 + |-1/3| + 1/9 + |-1/27| + 1/81 + … the geometric series with r = 1/3. This series converges, so the alternating series converges absolutely.

 

2. an= (-1)nn/(n2 + 2)

Answer

1. The series converges conditionally.

Solution

1. To test absolute convergence, test the series with terms |an| = n/(n2 + 2). The terms look like n/n2 = 1/n, which diverges. Because n/(n2 + 2) < 1/n, we cannot apply the Comparison Test. Instead, apply the Limit Comparison Test. limn → ∞ n/(n2 + 2)/(1/n)= limn → ∞ 1/(1 + 2/n2) = 1 Consequently, the series with terms |an| diverges. To test conditional convergence, we apply the Alternating Series Test. Consider the function f(x)= x/(x2 + 2) and compute f'(x)= (2 – x2)/(x2 + 2)2. Because f'(x) is negative for x > √2, the terms |an| are eventually decreasing. Also, limx → ∞ f(x)= 0, so limn → ∞,/sub> an= 0. Then by the Alternating Series Test, the series converges. Because it does not converge absolutely, the series converges conditionally.

 

3. an = (-1)n(n2 – 2n + 3)/(n2 + 2n + 5)

Answer

1. The series diverges.

Solution

1. The numerator and denominator have the same highest power of n, so apply the nth term test limn → ∞(n2 – 2n + 3)/(n2 + 2n + 5) = limn → ∞(1 – 2/n + 3/n2)/(1 + 2/n + 5/n2)= 1 Because the terms do not go to 0, the series diverges.

 

4. an= cos(n)/n2

Answer

1. The series converges absolutely.

Solution

1. Note |an| ≤ 1/n2, a p-series with p= 2, converging by the Integral Test. Because the series with term |an| converges, the original series converges absolutely.

 

5. For which p does the alternating p-series 1 – 1/2p + 1/3p – 1/4p + … converge absolutely, converge conditionally, or diverge?

Answer

1. Converges absolutely for p > 1, converges conditionally for 0 < p ≤ 1, and diverges for p ≤ 0.

Solution

1. To test absolute convergence, we test the series: 1 + |-1/2p| + |1/3p| + |-1/4p| + … the regular p-series. From the Integral Test, we know the p-series converges for p > 1 and diverges for p ≤ 1. Consequently, the alternating p-series converges absolutely for p > 1. For 0 < p ≤ 1, apply the Alternating Series Test. For f(x)= 1/xp, we find f'(x)= -p/xp+1 so f(x) is decreasing. Also, limn → ∞ 1/np= 0 so the alternating p-series converges. Because the series does not converge absolutely in this range of p-values, the series converges conditionally. For p ≤ 0, the series diverges by the nth term test.

Numerical Sequences and Series: The comparison test

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More Challenging Problems 

 

Determine if the series, a1 + a2 + a3 + a4 + … converges or diverges, where:

1. an= 1/(3n + 2)

Answer

1. Converges

Solution

1. First, observe that 0 < 1/(3n + 2) < 1/3n The series with bn = 1/3n is a geometric series with r= 1/3 and so converges. Then the series with an= 1/(3n + 2) converges by the comparison test.

 

2. an= 1/(n – 1)

Answer

1. Diverges

Solution

1. First, observe that 1/(n – 1) > 1/n The series with bn= 1/n is the harmonic series and so diverges. Then the series with an = 1/(n – 1) diverges by the comparison test.

 

3. an = 1/(n2 + n)

Answer

1. Converges

Solution

1. This is a little trickier, because we could compare this an with 1/n (the harmonic series, so diverges) or with 1/n2 (a p-series with p = 2 > 1, so converges). Which should be used?
Note: 1/(n2 + n) < 1/n and 1/(n2 + n) < 1/n2 Being less than a diverging series tells us nothing, but being less than a converging series tells us that the series of an converges.

 

4. an = 1/(3n – 2)

Answer

1. Converges

Solution

1. The Comparison Test cannot be applied, because 1/(3n – 2) > 1/3n and although the geometric series ∑ 1/3n converges, being greater than a converging series tells us nothing. We can use the Limit Comparison Test, because as n → ∞, (1/(3n – 2)/(1/3n) = (3n)/(3n – 2) = 1/(1 – 2/3n) → 1 Then by the Limit Comparison Test, ∑ 1/(3n – 2) converges because ∑ 1/3n converges.

 

5. an = (√(n3 + 4))/(n2 + 2n + 3)

Answer

1. Diverges

Solution

1. The first issue is to find what series to use for comparison. For large n the numerator √(n3 + 4) is close to n3/2. For large n the denominator n2 + 2n + 3 is close to n2. Then the fraction an is close to n3/2/n2 = 1/n1/2. This is the general term of the series we use for comparison. As n → ∞, (√(n3 + 4))/(n2 + 2n + 3) / (1/n1/2) = (n3/2 √(1 + 4/n3)) / (n2(1 + 2/n + 3/n2)) ⋅ n1/2 = (√(1 + 4/n3)) / (1 + 2/n + 3/n2) → 1 Being a p-series with p = 1/2 < 1, ∑1/n1/2 diverges, so the original series diverges by the Limit Comparison Test.

 

6. an = (3n + 2)/(5n + 4)

Answer

1. Converges

Solution

1. For large n the ratio (3n + 2)/(5n + 4) is very close to 3n/5n, the general term of a geometric series with ratio 3/5, hence converging. As n → ∞, ((3n + 2)/(5n + 4))/(3n/5n) = ((3n(1 + 2/3n))/(5n(1 + 4/5n))⋅(5n/3n) = (1 + 2/3n)/(1 + 4/5n) → 1 Then by the Limit Comparison Test, the original series converges because the geometric series converges.

 

 

Numerical Sequences and Series: The integral test

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More Challenging Problems

Determine if the series, a1 + a2 + a3 + a4 +… converges or diverges, when:

1. an = n/(n2 + 1)

Answer

1. Diverges

Solution

1. f(x)= x/(x2 + 1) is positive, continuous, and decreasing because f'(x)= (1 – x2)/(1 + x2)2 is negative for x > 1. Substituting u= x2 + 1 so du= 2x dx and x dx = (1/2)du, we see ∫x/(x2 + 1) dx = (1/2) ∫du/u = (1/2)ln(u) = (1/2)ln(x2 + 1) This diverges to ∞ as x → ∞, so the series diverges by the Integral Test.

 

2.an= n2/(n3 + 1)

Answer

1. Diverges

Solution

1. f(x)= x2/(x3 + 1) is positive, continuous. Now f'(x)= x(2 – x3)/(1 + x3)2 is negative for x > 21/3≈ 1.2599, so we can apply the Integral test only to the series starting with n= 2. This is fine: the n = 1 term cannot affect the convergence of the series. Substituting u= x3 + 1, so du = 3x2 dx and x2 dx = (1/3)du, we see ∫x2/(x3 + 1) dx = (1/3) ∫du/u = (1/3) ln(u) = (1/3) ln(x3 + 1) This diverges to ∞ as x → ∞, so the series diverges by the Integral Test.

 

3. an= n/(n2 + 1)2

Answer

1. Converges

Solution

1. f(x)= x/(x2 + 1)2 is positive, continuous, and decreasing because f'(x)= (1 – 3x2)/(1 + x2)3is negative for x > 1. Substituting u = x2 + 1 so du = 2x dx and x dx = (1/2)du, we see: ∫x/(x2 + 1)2 dx = (1/2)∫ du/u2 = -(1/2)(1/u) = -(1/2)(1/(x2 + 1)) This converges, so the series converges by the Integral Test.

 

4. an=1/(n ln(n)), starting with n = 2

Answer

1. Diverges

Solution

1. f(x)= 1/(x ln(x)) is positive, continuous, and decreasing because f'(x)= -(1 + ln(x))/(x2ln2(x)) is negative for > 1. Substituting u= ln(x) so du = (1/x)dx, we see ∫1/(x ln(x)) dx = ∫ du/u = ln(u) = ln(ln(x)) This diverges to ∞ as x → ∞, so the series diverges by the Integral Test.

 

Numerical Sequences and Series: Zeno’s Paradox

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More Challenging Problems

1. Find the sum of the series 1 – 1/2 + 1/4 – 1/8 + …

Answer

1. The series sums to 2/3.

Solution

 

2. Find the sum of the series 3 + 3/2 + 3/4 + 3/8 + …

Answer

1. The series sums to 6.

Solution

1. Factoring out a 3, this series is 3⋅(1 + 1/2 + 1/4 + 1/8 + …) The series in parentheses is a geometric series with ratio r = 1/2, converging because |r| < 1. This series sums to 1/(1 – r) = 1/(1 – (1/2)) = 2, so the original series sums to 3⋅2 = 6

 

3. Find the sum of the series 1/9 + 1/27 + 1/81 + 1/243 + …

Answer

1. The series sums to 1/6.

Solution

1. Factoring out a 1/9 from every term, we see the series is (1/9)⋅(1 + 1/3 + 1/9 + 1/27 + …) The series in parentheses is a geometric series with ratio r = 1/3, converging because |r| < 1. This series sums to 1/(1 – r) = 3/2, so the original series sums to (1/9)⋅(3/2) = 1/6

 

4. Find the sum of the series 1 + 2 + 4 + 8 + 16 + … + 128

Answer

1. The series sums to 255

Solution

1. Recall for all r, 1 + r + r2 + … + rN-1= (1 – rN)/(1 – r) Because 128= 27, we see that 1 + 2 + 4 + … + 128 = (1 – 28)/(1 – 2) = (1 – 256)/(1-2) = -255.

 

5. Generalize problem 4 to 1 + 2 + 4 + … + 2n

Answer

1. The series sums to 2n+1– 1

Solution

1. Recall for all r, 1 + r + r2 + … + rN-1= (1 – rN)/(1 – r) Then 1 + 2 + 4 + … + 2n= (1 – 2n+1)/(1 – 2)= 2n+1– 1.

 

Numerical Sequences and Series

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More Challenging Problems

In problems 1 – 4 determine if the sequences converge or diverge. For those that converge, find the limits.

 

1. an= (n2 + 3)/(2n2 – 5)

Answer

1. Converges to 1/2.

Solutions

2. Factoring n2 from numerator and denominator, we find limn→ ∞ (n2 + 3)/(2n2 – 5) = limn<→ ∞(1 + 3/n2)/(2 – 5/n2) = (1 + 0)/(2 – 0) = 1/2

 

2. an = (n2 + 3)/(2n3 – 5)

Answer

1. Converges to 0.

Solution

1. Factoring n3 from numerator and denominator, we find limn → ∞(n2 + 3)/(2n3 – 5)= limn → ∞(1/n + 3/n3)/(2 – 5/n3) = (0 + 0)/(2 – 0) = 0/2 = 0

 

3. an = (n2 + 3)/(2n – 5)

Answer

1. Diverges

Solution

1. Factoring n from numerator and denominator, we find limn → ∞ (n2 + 3)/(2n – 5)= limn → ∞(n + 3/n)/(2 – 5/n) =limn → ∞ n/2= ∞

 

4. an = √(n+1) – √n

Answer

1. Converges to 0

Solution

1. Multiply by (√(n+1) + √n)/(√(n+1) + √n), obtaining limn → ∞ √(n+1) – √n = limn → ∞ (√(n+1) – √n)(√(n+1) + √n)/(√(n+1) + √n) = limn → ∞ (n + 1 – n)/(√(n+1) + √n) = limn → ∞ 1/(√(n+1) + √n) = 0

 

In problems 5 – 8 determine if the series ∑n=1∞(a)n converge or diverge. For those that converge, find the limits. In problem 6, start the sum at n = 2.

5. an = n2/(2n2 + 1)

Answer

1. Diverges

Solution

1. Because limn → ∞ n2/(2n2 + 1)= limn → ∞ 1/(2 + 1/n2) = 1/(2 + 0) = 1/2 the series diverges by the 9th term test.

 

6. an = 2/(n2 – 1)

Answer

1. Converges to 3/2

Solution

1. Apply the method of partial fractions and conclude: 2/(n2 – 1) = 1/(n – 1) – 1/(n + 1) Starting the series from n = 2, the first few terms are (1/(2-1) – 1/(2+1)) + (1/(3-1) – 1/(3+1)) + (1/(4-1) – 1/(4+1)) + (1/(5-1) – 1/(5+1)) + … = (1 – 1/3) + (1/2 – 1/4) + (1/3 – 1/5) + (1/4 – 1/6) + … = 1 + 1/2 = 3/2 because 1/3 occurs twice, once + and once -, 1/4 occurs twice, once + and once -, and so on. Only the 1 and 1/2 are not cancelled by later terms.

 

7. an = n sin(1/n)

Answer

1. Diverges

Solution

1. Note that n sin(1/n) = sin(1/n)/(1/n). Because 1/n → 0 as n → ∞, and recalling limx → 0 sin(x)/x = 1, we see limn → ∞n sin(1/n) = 1. The series diverges by the nth term test.

 

8. an = arctan(n)

Answer

1. Diverges

Solution

1. As n → ∞, arctan(n) → π/2, so the series diverges by the nth term test.