**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.