Examples for Section 8.7
Alternating Series, Absolute and Conditional Convergence

Example

The series \(TraditionalForm
      \`∑\+\(n = 1\)\%∞\(\((\(-1\))\)\^\(n + 1\)\) 
          \((10\/n)\)\^n\) converges because:
1) It is alternating.
2) The terms are decreasing: \(TraditionalForm
      \`\((10\/n)\)\^n ≥ \((10\/\(n + 1\))\)\^n ≥ 
        \((10\/\(n + 1\))\)\^\(n + 1\)\) for n ≥ 10.
3) The terms approach 0: \(TraditionalForm
      \`lim\_\(n -> ∞\)\((10\/n)\)\^n = 
        \(lim\_\(n -> ∞\)e\^\(n(ln(10) - ln(n))\) = \ 
          \(e\^\(-∞\) = 0. \)\)\)

NSum[(((-1))^(n + 1)) ((((10/n)))^n), {n, 1, ∞}]

Example

The series \(TraditionalForm
      \`∑\+\(n = 1\)\%∞\(\((\(-1\))\)\^\(n + 1\)\) 
          \(sin(1\/n)\)\) converges because:
1) It is alternating.
2) The terms are decreasing: \(TraditionalForm
      \`1\/n ≥ 1\/\(n + 1\) => 
        sin(1\/n) ≥ sin(1\/\(n + 1\))\) since sin(x) is increasing for 0 < x < π/2.
3) The terms approach 0: \(TraditionalForm
      \`lim\_\(n -> ∞\)sin(1\/n) = 
        \(sin(lim\_\(n -> ∞\)1\/n) = \(sin(0) = 0. \)\)\)

NSum[(((-1))^(n + 1)) Sin[(1/n)], {n, 1, ∞}]

p. 661

11.

The series \(TraditionalForm
      \`∑\+\(n = 1\)\%∞\(\((\(-1\))\)\^\(n + 1\)\) 
          \((0.1)\)\^n\) is absolutely convergent because (∑_(n = 1)^∞)(((0.1))^n)  is a convergent geometric series. The original series is also a geometric series: \(TraditionalForm
      \`∑\+\(n = 1\)\%∞\(\((\(-1\))\)\^\(n + 1\)\) 
            \((0.1)\)\^n\  = \ 
        \(\(-\(∑\+\(n = 1\)\%∞\((\(-0.1\))\)\^n\)\) = 
          \(\(-\(\((\(-0.1\))\)\/\(1 - \((\(-0.1\))\)\)\)\) = 
            \(1\/11 = 0.090909 ... \)\)\)\)

NSum[(((-1))^(n + 1)) (((0.1))^n), {n, 1, ∞}]

21.

The series \(TraditionalForm
      \`∑\+\(n = 1\)\%∞\(\((\(-1\))\)\^\(n + 1\)\) 
          \((\(1 + n\)\/n\^2)\)\) is convergent because it is an alternating series whose terms are decreasing and approach 0: \(TraditionalForm
      \`lim\_\(n -> ∞\)\(1 + n\)\/n\^2 = 0\). The series is not absolutely convergent because the series \(TraditionalForm
      \`∑\+\(n = 1\)\%∞\((\(1 + n\)\/n\^2)\)\) diverges: ((1 + n)/(n^2)) = (1/(n^2)) + (1/n) > (1/n) and the harmonic series (∑_(n = 1)^∞) (1/n) diverges. Therefore, the original series is conditionally convergent.

NSum[
      (((-1))^(n + 1)) (((1 + n)/(n^2))), {n, 1, ∞}]

45.

Estimate the magnitude of the error in using the sum of the first four terms to approximate the sum of the series \(TraditionalForm
      \`∑\+\(n = 0\)\%∞\(\((\(-1\))\)\^n\) 1\/n\).

Since the series is alternating, the error is less than the absolute value of the 5th term: \(TraditionalForm
      \`\( | error | \( ≤ \((1\/5)\)\)\) = 0.2\).

1 - (1/2) + (1/3) - (1/4) // N

The series actually sums to ln(2).

Log[2] // N

The error in using the firts four terms is:

(1 - (1/2) + (1/3) - (1/4)) - Log[2] // N