Recitation 21

Example 1: Determine whether the series is absolutely convergent, conditionally convergent, or divergent. (a) \sum_{n=1}^\infty\frac{n}{5^n}; (b) \sum_{n=1}^\infty\frac{10^n}{(n+1)4^{2n+1}}; (c) \sum_{n=1}^\infty\frac{\cos(n\pi/3)}{n!}; (d) \sum_{n=1}^\infty(1+1/n)^{n^2}.

Hint: (a, b) Use the ratio test. (b) Compare the |\frac{\cos(n\pi/3)}{n!}| with \frac{1}{n!}. (d) Note that \lim_{n\to\infty}(1+1/n)^{n^2}=\infty.

Example 2: The terms of a series are defined recursively by the equations a_1 = 2, a_{n+1}=\frac{5n+1}{4n+3}a_n. Determine whether \sum a_n converges or diverges.

Hint: Note that the sequence is (ultimately) increasing and hence does not converge to 0.

Example 3: For what values of x does the series \sum_{n=1}^\infty\frac{(x-2)^n}{n^2+1} converge?

Hint: If x=2, the series converges apparently. Otherwise, observe that \lim_{n\to\infty}|a_{n+1}/a_n| = |x-2|. The ratio test shows that if 1<x<3 then the series converges and if x<1 or x>3 then the series diverges. If x = 1, then the series converges by the alternating series test. If x=3, then the series converges by the integral test. Therefore the series converges if and only if 1\leq x \leq 3.

Example 4: Test the series for convergence or divergence. (a) \sum_{n=1}^\infty (-1)^{n+1}\frac{n^2}{n^3+4}; (b) \sum_{n=1}^\infty(-1)^n\sin(\pi/n).

Hint: (a) Note that this is an alternating sequence, (n+1)^2/((n+1)^3+4) < n^2/(n^3+4) for all n > 1 and n^2/(n^3+4)\to 0 as n\to\infty. (b) Note that this is an alternating sequence, \sin(\pi/n+1) < \sin(\pi/n) for all n > 1 and \sin(\pi/n)\to 0 as n\to\infty.

Example 5: Show that the series \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^6} is convergent. How many terms of the series do we need to add in order to find the sum to the accuracy of 0.00005?

Solution: Note that this is an alternating series, 1/(n+1)^6 < 1/n^6 for all n and 1/n^6\to 0 as n\to\infty. By the alternating series test, the series converges. For alternating series, the error between s_n (the partial sum of the first n terms) and the sum of the series is bounded by a_{n+1}. In order to find the sum to the accuracy of 0.00005, it is sufficient to find n such that 1/(n+1)^6 < 0.00005, hence n = 5 suffices.

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