Math 181H: Hour Exam 3 (Solutions)

Prof. S. Smith: 28 April 1992

Problem 1: Test for convergence (and evaluate if possible) ∑_{n=0}^{∞} \frac{2^{n}−1}{3^{n}}.

Get ∑(\frac{2}{3})^{n} − ∑(\frac{1}{3})^{n} = \frac{1}{1− \frac{2}{3}} − \frac{1}{1− \frac{1}{3}} = 3 − \frac{3}{2} = \frac{3}{2}.

Problem 2: Test for convergence: ∑_{n=1}^{∞} \frac{1}{1+ln n}.

Can do comparison: for n ≥ 1 have n ≥ 1 + ln n, so \frac{1}{1+ln n} ≥ \frac{1}{n} = harmonic; diverges.

Problem 3: Test for convergence: ∑_{n=0}^{∞} \frac{n^{n}}{(3n+1)^{n}}.

Root:lim_{n→∞}(\frac{n^{n}}{(3n+1)^{n}})^{\frac{1}{n}} = lim_{n} \frac{n}{3n+1} = lim_{n} \frac{1}{3+ \frac{1}{n}} = \frac{1}{3} < 1, so converges.

Problem 4: Test for convergence: ∑_{n=0}^{∞} \frac{(n+3)!}{3!n!3^{n}}.

Ratio:lim_{n→∞} \frac{(n+4)!}{3!(n+1)!3^{n+1}} \frac{3!n!3^{n}}{(n+3)!} = lim_{n} \frac{n+4}{3(n+1)} = lim_{n} \frac{1+ \frac{4}{n}}{3(1+ \frac{1}{n})} = \frac{1}{3} < 1, so converges.

Problem 5: Determine the interval of convergence of ∑_{n=0}^{∞} \frac{x^{n}e^{n}}{n+1} .

Ratio: lim_{n→∞} \frac{x^{n+1}e^{n+1}}{n+2} \frac{n+1}{x^{n}e^{n}} = lim_{n} xe\frac{n+1}{n+2} = xe, so converges for |x| < \frac{1}{e}. At ±\frac{1}{e} get ∑_{n} \frac{(±1)^{n}}{n+1} . For +\frac{1}{e}: harmonic, diverges; for −\frac{1}{e}: alt. harmonic, converges. So interval is [−\frac{1}{e}, \frac{1}{e}).

Problem 6: Determine the (Maclaurin) series at a = 0 for cos \frac{x}{π}.

Derivs: cos \frac{x}{π}, −\frac{1}{π} sin \frac{x}{π}, −\frac{1}{π^{2}} cos \frac{x}{π}, \frac{1}{π^{3}} sin \frac{x}{π}, \frac{1}{π^{4}} cos \frac{x}{π}, …; at a = 0: 1, 0, −\frac{1}{π^{2}} , 0, \frac{1}{π^{4}} , . . .. So series is 1 − \frac{1}{π^{2}} \frac{x^{2}}{2!} + \frac{1}{π^{4}} \frac{x^{4}}{4!} − … = ∑_{n=0}^{∞} \frac{(−1)^{n}}{π^{2n}} \frac{x^{2n}}{(2n)!}.

Problem 7: Determine the Taylor series about a = −1 for \frac{1}{x}.

Derivs: \frac{1}{x}, −\frac{1}{x^{2}} , \frac{2}{x^{3}} , −\frac{6}{x^{4}} , …; values at a = −1 are −1, −1, −2, −6, . . .. So series is −1 − (x + 1) − 2\frac{(x+1)^{2}}{2!} − 6\frac{(x+1)^{3}}{3!} − … = − ∑_{n=0}^{∞}(x + 1)^{n}.

Problem 8: Determine the binomial series for (1 + x)^{\frac{3}{2}} .

1 + \frac{3}{2}x + \frac{3}{2} \frac{1}{2} \frac{x^{2}}{2!} + \frac{3}{2} \frac{1}{2} \frac{−1}{2} \frac{x^{3}}{3!} + … + \frac{3}{2} \frac{1}{2} \frac{−1}{2} … (\frac{3}{2} − n + 1)\frac{x^{n}}{n!} + …

Problem 9: Use a Taylor polynomial, based at a = 0 and up to x^{5}-term, to approximate the value of sin 1; also estimate the error.

Approximately: 1 − \frac{1^{3}}{3!} + \frac{1^{5}}{5!} = 1 − \frac{1}{6} + \frac{1}{120} = \frac{101}{120}. Error at most \frac{1}{7!} = \frac{1}{5040} (so good to at least 3 decimal places).