M344: Calculus III
Section 11.5
So far, the convergence tests we have seen only apply to series with positive terms. Naturally, we must ask
What happens when those terms are not necessarily positive?
In particular, we want to look at series whose terms alternate between positive and negative. These series are called Alternating series.
Motivating Example¶
Consider the so-called alternating harmonic series, $$ 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - \cdots = \sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{1}{n}. $$
In general, the $n^{th}$ term of an alternating series is given by
$$
a_n = (-1)^{n-1}b_n \quad \text{or} \quad a_n = (-1)^nb_n
$$
where $b_n$ is a positive number. (In fact, $ b_n = \lvert a_n \rvert$.)
The Alternating Series Test¶
If the alternating series
$$
\sum_{n=1}^{\infty} (-1)^{n-1}b_n = b_1 - b_2 + b_3 - b_4 +\cdots, \quad\text{where}\quad b_n > 0
$$
satisfies
- $b_{n+1} \le b_n$ for all $n$, and
- $\lim\limits_{n\to\infty} b_n = 0$,
then the series is convergent.
In other words, if the terms of an alternating series decrease toward 0 in absolute value, then the series converges. Before we prove the Alternating Series test, let's look at the following example. Consider the alternating series
$$
\sum_{n=1}^{\infty} (-1)^n\dfrac{1}{n} = -1 + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} - \cdots
$$
We can plot the areas and you can see that some of these areas are positive while the others are negative. So the partial sums will continue to oscillate. Since $b_n \to 0$, each area is geting smaller and smaller and we can see that the sum is converging to some number $s$. (In this example, the sum converges to $-\ln(2)$.)
Proof
First, let's consider the even partial sums: $$\displaystyle \begin{aligned} s_2 & = b_1 - b_2 \ge 0 & \quad \text{since}\quad b_2 \le b_1 \\ s_4 & = s_2 + (b_3 - b_4) \ge s_2 & \quad\text{since}\quad b_4\le b_3 \end{aligned} $$
In general, we have that
$$ s_{2n} = s_{2n-2} + (b_{2n-1} - b_{2n}) \ge s_{2n-2} \quad \text{since}\quad b_{2n} \le b_{2n-1} $$
Thus we have that
$$ 0 \le s_2 \le s_4 \le s_6 \le \cdots \le s_{2n} \le \cdots $$
However, $s_{2n}$ can also be written as
$$ s_{2n} = b_1 - (b_2-b_3) - (b_4-b_5) - \cdots - (b_{2n-2}-b_{2n-1}) - b_{2n} $$
Every term in parentheses is positive, so the partial sum $s_{2n} \le b_1$ for all $n$. This implies that the sequence $\{s_{2n}\}$ of even partial sums is increasing and bounded above. Therefore, by the Monotonic Sequence Theorem, is is convergent and we can say that
$$ \lim\limits_{n\to\infty} s_{2n} = s $$
for some $s$. Now, let's compute the limit of the odd partial sums:
$$ \displaystyle \begin{aligned} \lim\limits_{n\to\infty} s_{2n+1} & = \lim\limits_{n\to\infty} (s_{2n} + b_{2n+1})\\ & = \lim\limits_{n\to\infty} s_{2n} + \lim\limits_{n\to\infty} b_{2n+1} \end{aligned} $$
Since $b_n\to 0$ as $n\to\infty$, we have that the $\lim\limits_{n\to\infty} b_{2n+1} = 0$ and from before, we know that $\lim\limits_{n\to\infty} s_{2n} = s$. Therefore,
$$ \lim\limits_{n\to\infty} s_{2n+1} = s. $$
Going back to our previous example,
$$
\sum_{n=1}^{\infty} (-1)^n\dfrac{1}{n} = -1 + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} - \cdots
$$
We need to make sure this series satisfies the conditions of the Alternating Series Test (A.S.T.)
- $b_{n+1} < b_n$
- $\lim\limits_{n\to\infty} b_n = 0$.
Since $a_n = (-1)^n b_n$, $b_n$ term is only $\dfrac{1}{n}$. We know (1) is true because
$$
\dfrac{1}{n+1} < \dfrac{1}{n}.
$$
For (2), we get
$$
\lim\limits_{n\to\infty}\dfrac{1}{n} = 0.
$$
Therefore, since both conditions are satisfied, we get that this series is convergent.
1. Example¶
Determine where the series is convergent or divergent.
$$
\sum_{n=1}^{\infty} (-1)^n \frac{n^2}{n^2 + n + 1}
$$
Check your answer
The series is divergent.
Video solution
2. Example¶
Determine where the series is convergent or divergent.
$$
\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{3 + 5n}
$$
Check your answer
The series is convergent.
Video solution
3. Example¶
Determine where the series is convergent or divergent.
$$
\sum_{n=1}^{\infty} \frac{n\,\cos(n \pi)}{2^n}
$$
Check your answer
$\cos(n\pi) = (-1)^n$, so this series is alternating. The AST applies, and the series is convergent.
Video solution
Estimating Sums¶
Once again, we can approximate the total sum $s$ of a convergent series by using the partial sum $s_n$ but it is not much use unless we can estimate the accuracy of this approximation. The remainder $R_n = s - s_n$ will tell us about the error involved when using $s \approx s_n$.
Alternating Series Estimation Theorem¶
If $s = \sum (-1)^{n-1}b_n$, where $b_n > 0$, is the sum of an alternating series that satisfies
- $b_{n+1} \le b_n$
- $\lim\limits_{n\to\infty} b_n = 0$.
Then,
$$
\lvert R_n\rvert = \lvert s - s_n \rvert \le b_{n+1}
$$
In other words, the size of the error is smaller than $b_{n+1}$, which is the absolute value of the first neglected term.
5. Example¶
Find the sum of the series correct to three decimal places.
$$
\sum_{n=0}^{\infty}\dfrac{(-1)^n}{n!}
$$
Video solution
Discussion¶
Questions? You can ask them here.