Section 6: Alternating Series
An alternating series is one of the form $\sum_{n=1}^\infty (-1)^{n-1}a_n$ or $\sum_{n=1}^\infty (-1)^na_n$, where $a_n \geq 0$. For example, $\sum_{n=1}^\infty \dfrac{(-1)^n}{\sqrt{n}}$ and $\sum_{n=1}^\infty \dfrac{(-1)^{n-1}}{2^n}$ are alternating series. But $\sum_{n=1}^\infty \dfrac{\sin{(n)}}{n^2}$ is not alternating: although $\sin$ oscillates between -1 and 1, it does not do so in an alternating manner: $\sin{3}$ is positive and $\sin{4}$ is negative, but $\sin{5}$ is also negative.
The Alternating Series Test: If $\sum_{n=1}^\infty (-1)^{n-1}a_n$ is an alternating series with $0 \leq a_{n+1} \leq a_n$ for all $n$ and $\lim_{n \to \infty} a_n = 0$, then the series converges. In other words, if the sequence $a_n$ decreases to 0, then the series $\sum (-1)^{n-1}a_n$ converges.
Example: Let's take a look at the alternating harmonic series $\sum_{n=1}^\infty \dfrac{(-1)^{n-1}}{n}$. $1/n$ is a decreasing sequence and $\lim_{n \to \infty} \frac{1}{n} = 0$, so the series converges.
Mini-Quiz: Is $\sum_{n=1}^\infty \dfrac{(-1)^n\ln{(n)}}{n}$ convergent or divergent?
If we have a convergent alternating series, we can estimate the error in the approximation by a partial sum as follows. Let $s = \sum_{n=1}^\infty (-1)^{n-1}a_n, s_N = \sum_{i=1}^N (-1)^{i-1}a_i, R_N = s - s_N$. Then $|R_N| \leq a_{n+1}$. In other words, the exact value s lies between any two consecutive partial sums.
Example: Let's calculate $\sum_{n=1}^\infty \dfrac{(-1)^{n-1}\ln{(n)}}{n^2}$ to two decimal places. To show $\ln{(n)}/n^2$ is decreasing, let $f(x) = \frac{\ln{x}}{x^2}$. Then $f'(x) = \frac{x - 2x\ln{x}}{x^4}$, which is negative for $x > e^{1/2} \approx 1.64$. So $\frac{\ln{(n)}}{n^2}$ is decreasing for $n \geq 2$ (it's okay if it's only decreasing when we start the series at $2$ instead of $1$). Furthermore, L'Hospital's rule shows that $\lim_{x \to \infty} \frac{\ln{x}}{x^2} = 0$, which means $\lim_{n \to \infty} \frac{\ln{(n)}}{n^2} = 0$, so the series converges.
Now we know the remainder satisfies $|R_N| \leq \frac{\ln{(N+1)}}{(N+1)^2}$. We want this to be $< 0.005$. It's not easy to solve $\ln{(N+1)}/(N+1)^2 < 0.005$ for $n$. But since $\ln{(n)}/n^2$ is decreasing, so is $\ln{(n+1)}/(n+1)^2$. So let's plug in some values. $\ln{(15+1)}/(15+1)^2 \approx 0.01, \ln{(24+1)}/(24+1)^2 \approx 0.0051, \ln{(25+1)}/(25+1)^2 \approx 0.0048$. So we need to go out 25 terms for the value to be correct to two decimal places. That approximation yields $$\frac{\ln{1}}{1^2} - \frac{\ln{2}}{2^2} + \frac{\ln{3}}{3^2} - \frac{\ln{4}}{4^2} + \ldots + \frac{\ln{24}}{24^2} - \frac{\ln{25}}{25^2} \approx -0.11.$$
Mini-Quiz: Calculate $\sum_{n=0}^\infty \dfrac{(-1)^n}{n!}$ to three decimal places (recall that $0! = 1$).
Although we can't use the alternating series test to deal with a series like $\sum_{n=1}^\infty \dfrac{\sin{(n)}}{n^2}$, we can still find out if it's convergent or not. A series $\sum a_n$ is called absolutely convergent if $\sum |a_n|$ is convergent. If $\sum a_n$ is convergent, but not absolutely convergent, then $\sum a_n$ is conditionally convergent. One important fact is that any absolutely convergent series is also convergent.
Example: Let's take a look at $\sum_{n=1}^\infty \dfrac{\sin{(n)}}{n^2}$. The absolute value of the $n$th term is $|\sin{(n)}|/n^2$. Since $|\sin{(n)}| \leq 1$ and $\sum \frac{1}{n^2}$ converges, the comparison test tells us that $\sum_{n=1}^\infty \dfrac{\sin{(n)}}{n^2}$ is absolutely convergent and thus convergent.
Mini-Quiz: Is $\sum_{n=1}^\infty \dfrac{\cos{(e^{\pi/n^2})}}{n\sqrt{n}}$ convergent?