Category Archives: Numerical sequences and series – Challenging

More Challenging Problems: The ratio and root tests

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Introductory Problems 

 

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

 

1. an= (2⋅4⋅ … ⋅(2n))/(2n)!

Answer

1. The series converges.

Solution

    1. Factorials suggest the ratio test. The only trick is to simplify the numerator of an by                         factoring a 2 out of factor and grouping them together, so 2⋅4⋅ … ⋅(2n) = 2nn! Then after               simplification an+1/an= (2n+1/2n)⋅((n+1)!/(n!))⋅((2n)!/(2n+2)!) = (2(n+1)((2n)!))/((2n+2)(2n+1)                 ((2n)!)) = 1/(2n+1) This → 0 as n → ∞, so the series converges by the Ratio Test.

 

2. an= (nn)/(2n(2))

Answer

    1. The series converges.

Solution

    1. The presence of n in exponents suggests the Root Test. The trick is to recognize the                        denominator is 2n(2)= (2n)n, so an can be rewritten as (nn)/((2n)n)= (n/2n)n Then (an)1/n= n/2n.          Replacing n with x and applying l’Hopital’s rule, we see: limn → ∞(an)1/n = 0 so the series                converges by the Root Test.

 

3. an= ((3n)1)/(n1(2n)1)

Answer

    1. The series diverges.

Solution

    1.  Factorials suggest the ratio test. Grouping like factors together we find an+1/an=                               ((3n+3)!/(3n)!)⋅(n!/(n+1)!)⋅((2n)!/(2n+2)!) Canceling common factors gives an+1/an= ((3n+3)               (3n+2)(3n+1))/((n+1)(2n+1)(2n+2)) = (27n3 + quadratic and lower terms)/(4n3 + quadratic                 and lower terms) As n → ∞, the cubic terms dominate, so limn → ∞an+1/an= 27/4. The series           diverges by the Ratio Test.

 

 

More Challenging Problems: The alternating series test.

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Introductory Problems

In Problems 1 and 2, determine if the series converges absolutely, converges conditionally, or diverges.

 

1. an= (-1)nn/en

Answer

    1. The series converges absolutely.

Solution

    1. Take f(x)= x/ex, so an= (-1)nf(n). Note that f'(x)= (1 – x)/ex, so f(x) is decreasing for x ≥ 1, and           by l’Hopital’s rule limx → ∞x/ex= 0. Then by the Alternating Series Test, the series converges.           To test if the convergence is conditional or absolute consider the series bn= |an|. Apply the         Limit Comparison Test to bn and 1/n2: (n/en)/(1/n2)= n3/en → 0 as n → ∞ To see the last,                 replace n with x and apply l’Hopital’s rule three times. Then ∑bn converges by the Limit                 Comparison Test and so ∑an converges absolutely.

 

2. an= (-1)ncos(1/n)/n

Answer

    1. The series converges conditionally.

Solution

    1. Let f(x)= cos(1/x)/x, so an= (-1)nf(n). Note f'(x)= (-xcos(1/x) + sin(1/x))/x3, negative for x >                   1.163, so f(n) is decreasing for n > 1. Now limx → ∞cos(1/x)/x = 0, so by the Alternating Series         Test the series ∑an converges. To test if the convergence is conditional or absolute consider         the series bn= |an|. The series ∑bn diverges by the applying the Limit Comparison Test to             the harmonic series. Then the series ∑an converges conditionally.

 

3. Does the series: 1 + 1/3 – 1/2 – 1/4 + 1/5 + 1/7 – 1/6 – 1/8 + 1/9 + 1/11 – 1/10 – 1/12 + … converge absolutely?

Answer

    1. The series does not converge absolutely.

Solution

    1. If the series converged absolutely, then by the rearrangement theorem, every                                 rearrangement would converge to the same limit. But an obvious rearrangement of this               series is the alternating harmonic series, which does not converge absolutely.                                 Consequently, the original series does not converge absolutely.

 

 

More Challenging Problems: The comparison test

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Introductory Problems

 

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

1. an = sin(π/n)

Answer

    1. The series diverges.

Solution

    1. Recalling sin(x)/x → 1 as x → 0, we see sin(π/n)/(π/n) → 1 as n → ∞ Because ∑π/n is just π             times the harmonic series, which diverges, the given series diverges by the Limit                             Comparison Test.

 

2. an= 1/(1 + 2 + … + n)

Answer

    1. The series converges.

Solution

    1. Recall 1 + 2 + … + n = n(n+1)/2. (Using mathematical induction, this can be proved easliy.)             Note that 2/(n2 + n) < 2/n2, which is twice a convergent p-series, so the original series                     converges by the Comparison Test.

 

3. an = (1 + cos(n))/n3

Answer

    1. The series converges.

Solution

    1. Because -1 ≤ cos(n) ≤ 1, we see 0 ≤ 1 + cos(n) ≤ 2. Consequently, 0 ≤ (1 + cos(n))/n3 ≤ 2/n3.             The original series converges by comparison to a convergent p-series.

 

More Challenging Problems: Zeno’s paradox

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Introductory Problem

1. The whole shape pictured at the right has base and altitude 1. Find the area of the shaded red part remaining after the indicated squares are removed.

Answer

1. The area is 2/3.

Solution

1. The removed squares have side length 1/2, 1/4, 1/8, …, so the removed areas are: (1/2)2 + (1/4)2 + (1/8)2 + … = (1/2)2 + (1/22)2 + (1/23)2 + … = (1/22) + (1/22)2 + (1/22)3 + … = 1/4 + (1/4)2 + (1/4)3 +…This geometric series sums to (1/4)/(1 – (1/4)) = 1/3. Then the area of the shaded region is 1 – 1/3 = 2/3.

 

 

2. The whole shape pictured at the right has base and altitude 1. Find the area of the shaded red part remaining after the indicated triangles are removed.

Answer

1. The area of the triangles removed sums to 1/2, so the area remaining is 0.

Solution

1. We see 1 removed triangle with base = altitude = 1/2, three with base = altitude = 1/4, 9 with base = altitude = 1/8, and so on. The removed areas sum to (1/2)(1/2)(1/2) + 3(1/2)(1/4)(1/4) + 9(1/2)(1/8)(1/8) + …= (1/2)(1/4 + 3/42 + 9/43 + …)= (1/8)(1 + 3/4 + (3/4)2 + …) The geometric series sums to 1/(1 – 3/4) = 4, so the areas of the removed triangles sum to 1/2.The area of a triangle with base 1 and altitude 1 is 1/2, so the shaded region has area = 0.

 

 

3. Express the repeating decimal 1.1717171717… as a rational number.

Answer

1. The rational form of the repeating decimal 1.1717171717… is 116/99.

Solution

1. Write the repeating decimal 1.1717171717… as 1 + 17/100 + 17/1002 + 17/1003 + … = 1 + (17/100)(1 + 1/100 + 1/1002 + … = 1 + (17/100)(1/(1 – 1/100)) = 1 + 17/99 = 116/99

 

 

More Challenging Problems: Numerical Sequences and Series

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Introductory Problems

1. Suppose a1 = 1 and for all n > 2, an = (an-1 + 5)/2. Show the sequence an converges and find    the limit to which it converges.

Answer

1. The sequence is increasing and bounded above by 5, so it converges. The limit of the terms is 5

Solution

1. Each term after a1 is the midpoint of 5 and the previous term, so the sequence is increasing and bounded above by 5. It follows from the Monotone Convergence Theorem that this sequence converges. Now that we know the limit limn → ∞an exists, denote this limit by α. Taking the n → ∞ limit of both sides of the equation: an= (an-1 + 5)/2 we find: α= (α + 5)/2 Solving for α gives α= 5.

 

2. Suppose a1 = 1, a2 = 1, and for all n > 2, an= an-1 + an-2. Give numerical evidence that the sequence bn = an/an+1 converges. Assuming the sequence converges, find the limit to which it converges.

Answer

1. Convergence follows from two applications of the Monotone Convergence Theorem, and an estimate on the difference of successive terms. The sequence converges to (-1 + √5)/2.

Solution

1. The first few values of the ratios bn= an/an+1 are 1, .5, .6667, .6000, .6250, .6154, .6190, .6176, .6182, .6180, .6181, .6180, .6180, … . It appears that the sequence does converge. Now assuming the limit limn → ∞an/an+1 exists, denote this limit by α. Divide both sides of the equation an= an-1 + an-2 by an-1, obtaining an/an-1= 1 + an-2/an-1 Taking the n → ∞ limit of both sides of this equation gives 1/α= 1 + α Multiplying both sides by α gives a quadratic equation with solutions α= (-1 ± √5)/2. Being the limit of ratios of positive terms, α cannot be negative, so α= (-1 + √5)/2

 

3. Does the series ∑n=1∞an with an = log(n/(n+1)) converge or diverge?

Answer

1. The series diverges.

Solution

1. This is easy if you recall the quotient property of logarithms log(n/(n+1)) = log(n) – log(n+1) So the series telescopes (log(1) – log(2)) + (log(2) – log(3)) + (log(3) – log(4)) + … so the nth partial sum is Sn = log(1) – log(n+1) = -log(n+1), which diverges to -∞.

 

4. Suppose a1= 1/2 and for all n > 1, an= (1/2)n + an-1. Does the series a1 + a2 + a3 + … converge or diverge?

Answer

1. The series diverges.

Solution

1. Let’s look for a pattern in the first few terms: a1= 1/2 a2= 1/22 + 1/2 = (1 + 2)/22 a3= 1/23 + (1 + 2)/22 = (1 + 2 + 22)/23 and in general an = (1 + 2 + … + 2n)/2n+1 Now recall 1 + 2 + 22 + … + 2n= 2n+1 – 1. (Easy to prove by induction if you don’t recall this result.) So we see that: an= (2n+1 – 1)/2n+1 and the limn → ∞an= 1. The series diverges by the nth term test.

 

 

 

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.