Category Archives: More Challenging Problems

More Challenging Problems: Using power series to sum numerical series

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Introductory Problems

 

1. Sum the series ∑n=1∞((-1)nπ2n+1)/(42n+1(2n+1)!)

Answer

1. . .

Solution

1.

 

2. Sum the series π3/3 – π5/5⋅3! + π7/7⋅5! – π9/9⋅7! + … 

Answer

1. The series sums to π.

Solution

1. The number π likely is the point at which the series is evaluated, so the series is f(x)= x3/3 – x5/5⋅3! + x7/7⋅5! – x9/9⋅7! + … So the series to sum is f(π). The x3/3, x5/5, x7/7, x9/9, … suggests an integral, so let’s differentiate the series f'(x)= x2 – x4/3! + x6/5! – x8/7! + … The denominators and alternating signs suggest sin(x), but the exponents are wrong. Try factoring out an x f'(x)= x(x – x3/3! + x5/5! – x7/7! + …) x sin(x) Then f(x)= ∫f'(x) dx= ∫x sin(x) dx= -x cos(x) + sin(x), using integration by parts. Then the series sums to f(π)= -π cos(π) + sin(π) = π.

 

3. Sum the series 2 + 1/3 + 2/34 + 1/35 + 2/38 + 1/39 + 2/312 + 1/313 + … 

Answer

1. The series sums to 189/80.

Solution

1. This appears to be the series: 2 + x + 2x4 + x5 + 2x8 + x9 + 2x12 + x13 + …  evaluated at x= 1/3. Segregate the terms with multiples of 2 from those without 2 + x + 2x4 + x5 + 2x8 + x9 + 2x12 + x13 + … = (2 + 2x4 + 2x8 + 2x12 + …) + (x + x5 + x9 + x13 + …) Factor out a 2 from the first, a x from the second, obtaining 2(1 + x4 + x8 + x12 + … ) + x(1 + x4 + x8 + x12 + … ) = (2 + x)(1 + x4 + x8 + x12 + … ) = (2 + x)/(1 – x4) At x= 1/3, the series sums to (2 + 1/3)/(1 – 1/34) = 189/80.

 

4. Suppose a1 = 1 and for all n > 1, an = (1/2)n + (an-1/2). Find the sum of the series a1 + a2 + a3 +… 

Answer

1. The series sums to 3.

Solution

1. First, try to find an expression for an in terms of n, not an-1.                                                               a2 = (1/2)2 + a1/2 = 1/22 + 1/2 = 3/22                                                                                                              a3 = (1/2)3 + a2/2 = 1/23 + 3/23 = 4/23                                                                                                            a4 = (1/2)4 + a3/2 = 1/24 + 4/24 = 5/24                                                                                                            a5 = (1/2)5 + a4/2 = 1/25 + 5/25 = 6/25                                                                                                        and in general an = (n+1)/2n for n ≥ 1. That is, we want to sum the series :                                         2/21 + 3/22 + 4/23 + 5/24 + 6/25 + …                                                                                                           Next, recall the geometric series:  1/(1 – x)= 1 + x + x2 + x3 + x4 + … . Differentiating gives: 1/(1 – x)2= 1 + 2x + 3x2 + 4x3 + 5x4 + … Except for the initial 1, when evaluated at x= 1/2 this is the series we want to sum. That is, 1/(1 – 1/2)2 – 1 = 2/21 + 3/22 + 4/23 + 5/24 + 6/25 + …= 3

 

 

More challenging problems: Differentiating and integrating power series

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Introductory Problems

1. Find a power series y(x) = ∑n = 0∞anxn that satisfies y’ = y and y(0) = 1.

Answer

1. 1. The series is y(x) = 1 + x + x2/2! + x3/3! + … = ex.

Solution

 

2. Find a power series y(x) = ∑n = 0∞anxn that satisfies xy” + (1 – x)y’ – y = 0 and y(0) = 2.

Answer

1.  The series is y(x) = 2 + 2x + x2 + x3/3 + x4/(4⋅3) + x5/(5⋅4⋅3) +…

Solution

1. Differentiating y(x) = a0 + a1x + a2x2 + a3x3 + a4x4 + … gives:                                                                y'(x) = a1 + a22x + a33x2 + a44x3 + a55x4 + …y”(x) = a22 + a33⋅2x + a44⋅3x2 + a55⋅4x3 + a66⋅5x4 +   Then:  xy'(x) = a1x + a22x2 + a33x3 + a44x4 + a55x5 + …xy”(x) = a22x + a33⋅2x2 + a44⋅3x3 + a55⋅4x4 + a66⋅5x5 + … Equating coefficients of like powers of x gives:                                                           xy”           y’ – xy’          -y         = 0                     gives                                                                           x0          0             a1 – 0          -a0        = 0                     a1 = a0                                                                      x1       2a2         2a2 – a1          -a1        = 0                   4a2 – 2a1 = 0, so a2 = a1/2                                         x2    3⋅2a3        3a3 – 2a2        -a2        = 0                   9a3 – 3a2 = 0, so a3 = a2/3                                         x3    4⋅3a4        4a4 – 3a3        -a3        = 0                 16a4 – 4a3 = 0, so a4 = a3/4

 

All the coefficients can be expressed in terms of a0, giving an = a0/n! and consequently:

y(x) = a0(1 + x + x2/2 + x3/3! + x4/4! + …) = a0ex.The condition y(0) = 2 implies a0 = 2.

 

3. Prove that on its domain, (1 – sin(x))-2 = 1 + 2sin(x) + 3sin2(x) + 4sin3(x) + …

Solution

 

4. By dividing the series, find the coefficients, through the cubic term, of (1 – x)/ex.

Answer

1. The series for (1 – x)/ex begins 1 – 2x + (3/2)x2 – (2/3)x3.

Solution

1. Mimic long division of polynomials.

 

 

More Challenging Problems: Taylor’s theorem

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Introductory Problems

1. Find the Taylor series for ln(1 + x) expanded about x = 0. Find the radius of convergence of this series.

Answer

    1. The series is x – x2/2 + x3/3 – x4/4 + … . The radius of convergence is 1.

Solution

     1. To find the taylor series, compute the first few derivatives of f(x) = ln(1 + x) and look for patterns. f'(x)= 1/(1 + x) f”(x)= -1/(1 + x)2 f”'(x)= 2/(1 + x)3 f(4)(x)= -3!/(1 + x)4 and  in general f(n)(x)= (-1)n-1(n-1)!/(1 + x)n . Then f(n)(0)= (-1)n-1(n-1)! and so f(n)(0)/n! = (-1)n-1/n. Recalling f(0)= 0, the Taylor series is ln(1 + x)= x – x2/2 + x3/3 – x4/4 + … + (-1)n-1xn/n + … Apply the Ratio Test to find the radius of convergence limn → ∞|((-1)nxn+1/(n+1)) / ((-1)n-1xn/n)| = limn → ∞|(n/(n+1)) x| = |x| The series converges for |x| < 1, so the radius of convergence is 1.

 

2. Without computing any derivatives, find the Taylor series of cos(√(x2 + 1)) expanded about x = 0. 

Answer

1. The series is 1 – (x2 + 1)/2! + (x2 + 1)2/4! – (x2 + 1)3/6! + … + (-1)n(x2 + 1)n/(2n)! + …

Solution

1. Recall the Taylor series for cos(x): cos(x) = 1 – x2/2! + x4/4! – x6/6! + … + (-1)nx2n/(2n)! + … Substitute √(x2 + 1) for x, obtaining 1 – (x2 + 1)/2! + (x2 + 1)2/4! – (x2 + 1)3/6! + … + (-1)n(x2 + 1)n/(2n)! + …

 

3. Evaluate the limit limx → 0(tan(x) – sin(x))/x3 

Answer

1. The limit is 1/2.

Solution

1. Because the denominator is x3, we need the series for the numerator through the cubic term. The series for sin(x) is familiar: sin(x) = x – x3/3! + … . To find the terms of the Taylor series for tan(x) up through the cubic term, first compute the derivatives (tan(x))’= sec2(x) (tan(x))”= 2 sec(x) sec(x) tan(x) = 2 sec2(x) tan(x) (tan(x))”’= 2(2 sec(x) sec(x) tan(x))tan(x) + 2 sec2(x) sec2(x) = 4sec2(x)tan2(x) + 2sec4(x) Evaluating these derivatives at x = 0 gives tan'(0)= 1 tan”(0)= 0 tan”'(0)= 2 Then the first few terms of the Taylor series for tan(x) are x + x3/3 Up through cubic terms, tan(x) – sin(x) = (1/3 – (-1/3!))x3 = x3/2 Then the limit is 1/2.

 

4. For what value of k is this limit limx → 0(1 – kx2 – cos(3x))/x4 finite? Find the limit for that k.

Answer

1. The value of k is 9/4. For that value of k, the limit is 27/8.

Solution

1. The Taylor series for cos(3x) is 1 – (3x)2/2! + (3x)4/4! – (3x)6/6! + … . Because the denominator is x4, in order for the limit to exist the kx2 term must subtract out of the numerator, so k = 32/2! = 9/2. For k = 9/2, the value of the limit is the ratio of the coefficients of the quartic terms, so the limit is 34/4! = 81/24 = 27/8.

 

More Challenging Problems: The radius and interval of convergence

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Introductory Problems

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

1. ∑n=1∞ n! xn/(2n)!

Answer

1. The radius of convergence is ∞.

Solution

1. Factorials suggest the Ratio test. limn → ∞|((n+1)!xn+1/(2n+2)!) / (n!xn/(2n)!)| = limn → ∞|((n+1)!/n!)⋅((2n)!/(2n+2)!)x| = limn → ∞(n+1)/((2n+2)(2n+1))|x| = 0 for all x. This power series converges for all x, so the radius of convergence is ∞.

 

2. ∑n=1∞ nnxn/2n

Answer

1. The radius of convergence is 0. The interval of convergence is [0].

Solution

1. Exponents suggest the Root Test. limn → ∞(nn|x|n/2n)1/n= limn → ∞(n/2)|x| = ∞ for all x ≠ 0. Then the radius of convergence is 0, and the inverval of convergence is [0].

 

3. ∑n=1∞ n2xn/2n

Answer

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

Solution

1. The exponent of n2 suggests the Root Test will not be effective. Try the Ratio Test. limn → ∞|((n+1)2xn+1/2n+1) / (n2xn/2n)| = limn → ∞((n+1)2/n2)⋅(2n/2n+1)|x| = |x|/2. The interval of convergence includes |x|/2 < 1, that is, (-2,2), and so the radius of convergence is 2. To find the interval of convergence, test the endpoints of (-2,2). For x = 2 the series becomes ∑n=1∞ n22n/2n = ∑n=1∞ n2. This diverges by the nth term test. For x = -2 the series becomes ∑n=1∞ n2(-2)n/2n = ∑n=1∞ (-1)nn2. This diverges by the nth term test. Then the interval of convergence is (-2,2).

 

4. ∑n=1∞ (2x – 1)2n+1/n1/2

Answer

1. The radius of convergence is 1/2. The interval of convergence is (0, 1).

Solution

1. The exponent of n1/2 suggests the Root Test will not be effective. Try the Ratio Test. limn → ∞|((2x – 1)2n+3/(n+1)1/2) / ((2x – 1)2n+1/n1/2)| = limn → ∞(n/(n+1))1/2|(2x – 1)2| = |(2x – 1)2| The interval of convergence includes |(2x – 1)2| < 1, that is, |2x – 1| < 1, which gives 0 < x < 1. Then the radius of convergence is 1/2. To find the interval of convergence, test the endpoints of (0,1). For x = 1 the series becomes ∑n=1∞ 12n+1/n1/2 = ∑n=1∞ 1/n1/2, a diverging p-series. For x = 0 the series becomes ∑n=1∞ (-1)2n+1/n1/2 = -∑n=1∞ 1/n1/2, a diverging p-series. Then the interval of convergence is (0,1).

 

 

More Challenging Problems: Geometric series

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Introductory Problems

1. Write 1 + 2x + 4x2 + 8x3 + 16x4 + … as a ratio of polynomials.

Answer

1. The series sums to 1/(1 – 2x).

Solution

1. Rewrite the terms of the series 1 + 2x + 4×2 + 8×3 + 16×4 + … = 1 + 2x + (2x)2 + (2x)3 + (2x)4 + …So we see this series converges to 1/(1 – 2x).

 

2. Write x + x2 + x3 + x4 + … as a ratio of polynomials.

Answer

1. The series sums to x/(1 – x).

Solution

1. The series is the geometric series for 1/(1 – x), missing the 1, so x + x2 + x3 + x4 + … = (1 + x + x2 + x3 + x4 + … ) – 1= 1/(1 – x) – 1 = x/(1 – x).

 

3. Write 2 + x2 + x3 + x4 + 2x6 + x8 + x9 + x10 + 2x12 + x14 + x15 + x16 + 2x18 + … as a ratio of polynomials.

Answer

1. The series sums to (2 – x2 – x3)/(1 – x2 – x3 + x5).

Solution

1. The terms with coefficient 2 are x0, x6, x12, x18, … , in general, x6n. This suggests rearranging the series into terms of the form x2n and terms of the form x3n, obtaining:                                         1 + x2 + x4 + x6 + x8 + x10 + x12 + x14 + x16 + …                                                                                           +1 + x3 + x6 + x9 + x12 + x15 + x18 + …= 1/(1 – x2) + 1/(1 – x3)(2 – x2 – x3)/(1 – x2 – x3 + x5)

 

4. Write 1 + 2x + 3x2 + 4x3 + 5x4 + … as a ratio of polynomials.

Answer

1. The series converges to 1/(1 – 2x + x2).

Solution

1. First observe the sum 1 + 2x + 3x2 + 4x3 + 5x4 +… can be reorganized:1 + x + x2 + x3 + x4 +…= 1/(1 – x) x + x2 + x3 + x4 +…                                                                                                                            = x/(1 – x) x2 + x3 + x4 +…                                                                                                                               = x2/(1 – x) x3 + x4 +…                                                                                                                                     = x3/(1 – x) x4 +…                                                                                                                                               = x4/(1 – x)                                                                                                                                                           Consequently,  1 + 2x + 3x2 + 4x3 + 5x4 + …                                                                                                = (1 + x + x2 + x3 + …)/(1 – x)                                                                                                                            = 1/(1 – x)2                                                                                                                                                         = 1/(1 – 2x + x2)

 

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