Calculus II, Spring 2009 (http://www.math.nthu.edu.tw/˜ wangwc/)

Homework Assignment for Week 03

Assigned Mar 12, 2009.

1. Section 9.6: Problems 11, 15, 21, 23, 27, 28, 35, 39, 41, 44, 47.

2. Section 9.7: Problems 1, 3, 7, 15, 19, 25, 29, 33(show that equality holds), 35, 47(a), 50, 57, 58.

3. Find a power series whose interval of convergence is [1, 3). Do the same for (1, 3), [1, 3] and (1, 3], respectively.

4. (Challenge of the week)

Find the interval of convergence for the power series

1 + \frac{1}{2}x + \frac{1 · 3}{2 · 4}x^{2} + \frac{1 · 3 · 5}{2 · 4 · 6}x^{3} + ⋯ + \frac{1 ⋯ (2n − 1)}{2 ⋯ 2n}x^{n} + · · · .

Hint: Denote by a_{n} = \frac{1···(2n−1)}{2···2n} and define b_{2} = \frac{2}{3}, b_{3} = \frac{2·4}{3·5}, ⋯, b_{n} = \frac{2···(2n−2)}{3···(2n−1)}. For x = 1, the fact that b_{n} < 1 will help. For x = −1, compare a_{n} with b_{n} and 2b_{n} to conclude that a_{n} ∼ n^{−p}. What is p? Do the same for

1 + \frac{1}{3}x + \frac{1 · 4}{3 · 6}x^{2} + \frac{1 · 4 · 7}{3 · 6 · 9}x^{3} + ⋯ + \frac{1 ⋯ (3n − 2)}{3 ⋯ 3n}x^{n} + · · · .