Category Archives: Using power series to sum numerical series

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