Notes PDF
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.