Section 5.6

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1. (a) The series converges absolutely (p -series with p = 3 ), and so converges.

(c) The series diverges by the n th term for divergence since

\displaystyle{\lim_{n \to \infty}\frac{3n^2 - 1}{4n^2 + 2} = \frac{3}{4}}.

(e) The series does not converge absolutely (limit comparison test with a p -series with p = \dfrac{1}{2} ), but does converge conditionally by Leibniz’s theorem. Hence the series converges.

(g) Since \displaystyle{\lim_{n \to \infty}\frac{n!}{2^n} = \infty}, the terms \dfrac{(-1)^n n!}{2^n} do not have a limit as n \rightarrow \infty . Hence the series diverges by the n th term test for divergence.

2. (a) The series converges absolutely (limit comparison with a p -series with p = 3 ), and so converges.

(c) The series converges absolutely (ratio test, \rho = 0 ), and so converges.

(e) The series converges absolutely (ratio test, \rho = 0 ), and so converges.

(g) Since \displaystyle{\lim_{n \to \infty}\frac{n + 1}{2n - 1} = \frac{1}{2}}, the terms \dfrac{(-1)^n (n + 1)}{2n - 1} do not have a limit as n \rightarrow \infty . Hence the series diverges by the the n th term test for divergence.

3. (a) \displaystyle{s_{15} = \frac{34,361,893,981}{93,405,312,000} =  0.367879441171397}, to 15 decimal places

(b) |s - s_{15}| \le \dfrac{1}{16!} = 4.779 \times 10^{-14} , rounded to 17 decimal places

(c) s_9 = \dfrac{16,687}{45,360} = 0.367879 to 6 decimal places

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