Author Archives: mh225

Power Series: The radius and interval of convergence

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More Challenging Problems

Find the radius and interval of convergence of these power series.

1. ∑n=1∞xn/√n

Answer

1. The radius of convergence is 1, the interval of convergence is [-1,1).

Solution

1. Apply the Ratio Test limn → ∞|(xn+1/√(n+1))/(xn/√n)| = limn → ∞ |x|⋅√(n/(n+1)) = |x| By the Ratio Test, the series converges for |x| < 1, that is, -1 < x < 1. The radius of convergence is 1. At the left endpoint, the series becomes ∑n=1∞(-1)n/√n. This converges by the Alternating Series Test. At the right endpoint, the series becomes ∑n=1∞1/√n. This is a p-series with p = 1/2, so diverges. The interval of convergence is [-1,1).

 

2. ∑n=1∞(x-1)n/(n 5n)

Answer

1. The radius of convergence is 5, the interval of convergence is [-4,6).

Solution

1. Apply the Ratio Test limn → ∞|((x-1)n+1/((n+1)5n+1) / ((x-1)n/(n 5n)| = limn → ∞ |(x-1)n+1/(x-1)n|⋅(n/(n+1))⋅(5n/5n+1) = |x-1|/5 By the Ratio Test, the series converges for |x-1|/5 < 1, that is for -5 < x-1 < 5, or -4 < x < 6. The radius of convergence is 5. At the left endpoint the series becomes ∑n=1∞(-5)n/(n 5n) = ∑n=1∞(-1)n/n convergent by the Alternating Series Test. At the right endpoint the series becomes ∑n=1∞(5)n/(n 5n)= ∑n=1∞ 1/n This is the harmonic series, hence divergent. The interval of convergence is [-4, 6).

 

3. ∑n=2∞(-1)nxn/(2nln(n))

Answer

1. The radius of convergence is 2, the interval of convergence is (-2,2].

Solution

1. Apply the Ratio Test limn → ∞|((-1)n+1xn+1)/(2n+1ln(n+1)) / ((-1)nxn)/2nln(n))| = limn → ∞ |-(xn+1/xn)|⋅(2n/2n+1)⋅(ln(n)/ln(n+1)) = |-x|/2 where limn → ∞ln(n)/ln(n+1) = 1 is obtained by applying l’Hôpital’s rule to limx → ∞ln(x)/ln(x+1) By the Ratio Test, the series converges for |-x|/2 < 1, that is, for -2 < x < 2. The radius of convergence is 2. At the left endpoint, the series becomes ∑n=2∞(-1)n(-2)n/(2nln(n))= ∑n=2∞ 2n/(2n ln(n))) = ∑n=2∞1/ln(n) Now ln(n) < n, so 1/ln(n) > 1/n and the series diverges by comparison with the harmonic series. At the right endpoint, the series becomes ∑n=2∞(-1)n2n/(2nln(n)) = ∑n=2∞ (-1)n/ln(n) converging by the Alternating Series Test. The interval of convergence is (-2,2].

 

4. ∑n=1∞(3x + 1)n/n2

Answer

1. The radius of convergence is 1/3, the interval of convergence is [-2/3, 0].

Solution

1. Apply the Ratio Test limn → ∞|((3x+1)n+1/(n+1)2) / ((3x+1)n/n2)|= limn → ∞ |(3x+1)n+1/(3x+1)n|⋅(n2/(n+1)2)= |3x+1| By the Ratio Test, the series converges for |3x+1| < 1, that is, -1 < 3x+1 < 1, so -2 < 3x < 0, giving -2/3 < x < 0. The radius of convergence is 1/3. At the left endpoint, the series becomes ∑n=1∞ (-1)n/n2 convergent by the Alternating Series Test. At the right endpoint, the series becomes ∑n=1∞ 1n/n2 convergent, being a p-series with p= 2. The interval of convergence is [-2/3, 0].

 

5. ∑n=1∞xn/nn

Answer

1. The radius of convergence is ∞, the interval of convergence is (-∞, ∞).

Solution

1. Apply the Ratio Test: limn → ∞|(xn+1/(n+1)n+1)/(xn/nn)| = limn → ∞ |xn+1/xn|⋅((nn)/((n+1)n+1)) = limn → ∞|x|⋅(n/(n+1))n⋅(1/(n+1)) = 0 for all x Here we have used limn → ∞(1 + 1/n)n= e, we see limn → ∞(n/(n+1))n= 1/e to be sure we needn’t be concerned with the factor (n/(n+1))n in the limit. By the Ratio Test, this series converges for al x, so the radius of convergence is ∞ and the interval of convergence is (-∞, ∞).

 

 

Power Series: Geometric series

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More Challenging Problems

1. Find a power series for 1/(x2 – 1)

Answer

1. The series is -1 – x2 – x4 – x6 – x8-…

Solution

1. First, note that 1/(x2 – 1)= -1/(1 – x2). Now apply the series of Example 3 to conclude 1/(x2 – 1)= -(1 + x2 + x4 + x6 + x8 + …)

 

2. From the power series for 1/(x + 1) and for 1/(x – 1), use partial fractions to find a power series for 1/(x2 – 1). What assumption are you making in this approach?

Answer

    1. Adding the series for (1/2)(1/(x+1)) and for (1/2)(1/(x-1)), and rearranging the terms, we           obtain the series from problem 1.

Solution

    1. From Example 1 we know a series for 1/(1 + x) is 1 – x + x2 – x3 + x4– … The series for 1/(x – 1)= -(1/(1 – x)) is -1 – x – x2 – x3 – x4 – … Next, note that using a partial fractions expansion we have 1/(x2 – 1) = (1/2)(1/(x-1)) – (1/2)(1/(x+1)) Assuming we can add the series and rearrange them in any way we wish (This will be addressed in the next section), we find 1/(x2 – 1)= (1/2)(-1 – x – x2 – x3 – x4 – …) – (1/2)(1 – x + x2 – x3 + x4 – …) = -1 – x2 – x4; – x2 – … agreeing with       the solution to problem 1.

 

3. Find a power series for 1/(x2 + 5x + 6). On what interval is this series valid?

Answer

    1. The series is (1/2 – 1/3) – (1/4 – 1/9)x + (1/8 – 1/27)x2 – (1/16 – 1/81)x3 + …This series is valid                for |x| < 2.

Solution

    1. Using the method of partial fractions, we find: 1/(x2 + 5x + 6) = 1/((x + 2)(x + 3)) = 1/(x + 2) – 1/(x + 3). Using the method of Example 4 we find these power series: 1/(2 + x) = 1/(2(1 + x/2)) = (1/2)(1 – x/2 + x2/4 – x3/8 + x4/16 – …) 1/(3 + x) = 1/(3(1 + x/3)) = (1/3) (1 – x/3 + x2/9 – x3/27 + x4/81 – …) Subtracting these we find the series for 1/(x2 + 5x + 6) is: (1/2 – 1/3) – (1/4 –         1/9)x + (1/8 – 1/27)x2 – (1/16 – 1/81)x3 + … The series for 1/(2 + x) is valid for: |x/2| < 1, that is, |x| < 2. The series for 1/(3 + x) is valid for |x/3| < 1, that is, |x| < 3. Both expansions are valid for the intersection of these intervals, |x| < 2.

 

4. Find a power series for (2x + 5)/(x2 + 5x + 6). On what interval is this series valid?

Answer

    1. The series is (1/2 + 1/3) – (1/4 + 1/9)x + (1/8 + 1/27)x2 – (1/16 + 1/81)x3 + …This expansion is            valid for |x| < 2.

Solution

    1. Using the method of partial fractions, we find (2x + 5)/(x2 + 5x + 6) = (2x + 5)/((x + 2)(x + 3))= 1/(x + 2) + 1/(x + 3) Recalling the series for 1/(2 + x) and 1/(3 + x) from problem 3, we find the series for (2x + 5)/(x2 + 5x + 6) is the sum of these (1/2 + 1/3) – (1/4 + 1/9)x + (1/8 + 1/27)x2 – (1/16 + 1/81)x3 + … As in problem 3, this expansion is valid for |x| < 2.

 

5. Using -1/x = 1/(1 – (x+1)), find a power series for -1/x. On what interval is this series valid?

Answer

    1. The series is 1 + (x + 1) + (x + 1)2 + (x + 1)3 + (x + 1)4 + … The series is valid for -2 < x < 0.

Solution

    1. The expansion for 1/(1 – (x + 1)) is 1 + (x + 1) + (x + 1)2 + (x + 1)3 + (x + 1)4 + … This converges          for |x + 1| < 1. That is, -1 < x + 1 < 1, or -2 < x < 0

 

 

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: 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.

 

 

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.

 

 

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.