(ST5) Alternating Series Test#

In this lesson we are going to see how to:

  • use the Alternating Seriest Test to show that an alternating series is convergent.

Review Video#

Alternating Series#

An alternating series is a series that alternates between positive and negative terms:

\[ \sum_{n=1}^{\infty} (-1)^{n-1} b_n = b_1 - b_2 + b_3 - b_4 + \cdots \]

For example:

\[\begin{split} \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n} = -&1+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+\cdots \\[10pt] \sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n^2} = &1-\dfrac{1}{4}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{25}+\cdots \\[10pt] \end{split}\]

Usually, this is a result of there being a term like:

\[ (-1)^{n-1}\qquad (-1)^n \qquad (-1)^{n+1} \]

Warning

The Integral and Comparison Tests cannot be used when investigating the convergence of an alternating series, since those tests require positive terms.

The Test#

Alternating Series Test

Given an alternating series of the form:

\[ \sum_{n=1}^{\infty} (-1)^{n-1} b_n \qquad \text{where} \quad b_n>0 \]

If the \(b_n\) terms satisfy both:

\[ b_{n+1}\leq b_{n} \quad \text{for all } n \qquad \textbf{and} \qquad \lim_{n\to\infty}b_n = 0 \]

Then the series is convergent.

To apply this test we need to show:

  1. The sequence of \(b_n\) terms is decreasing. (Using inequalities or derivatives.)

  2. The sequence of \(b_n\) approaches \(0\).

Tip

  1. We can relax the condition on \(b_n\), so that \(b_{n+1}\leq b_{n}\) is eventually satisfied for all \(n\geq N\).

  2. This test also works if the \((-1)^{n-1}\) term is replaced by a similar alternating term.

Example 1#

Determine if the following series is convergent or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n} \]

(Click to see the steps.)

We see the alternating term \((-1)^n\) in our series, so we suspect the Alternating Series Test might be applicable. We begin by separating off this alternating piece, and writing the series as:

\[ \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n} = \sum_{n=1}^{\infty} (-1)^n\cdot \dfrac{1}{n} \]

This helps us identify the \(b_n\) term:

\[ b_n = \dfrac{1}{n} \]

The first condition we will show is that the limit of \(b_n\) is \(0\).

\[ \lim_{n\to\infty}b_n = \lim_{n\to\infty} \dfrac{1}{n} =0 \]

Since this is one of our \(\tfrac{c}{\infty}\)-forms.

Next, we show that the sequence \(\{b_n\}\) is decreasing, or more specifically that \(b_{n+1}\leq b_n\) for all \(n\geq 1\). To do this we will start by writing out the individual \(b_n\) and \(b_{n+1}\) terms:

\[ b_n = \dfrac{1}{n} \qquad \text{and}\qquad b_{n+1}=\dfrac{1}{n+1} \]

To actually prove the inequality we start with what we want and simplify until we get a statement that is clearly true:

\[\begin{split} & b_{n+1} &\leq b_n \\[10pt] \iff & \dfrac{1}{n+1} & \leq \dfrac{1}{n} \\[10pt] \iff & n & \leq n+1 \\[10pt] \end{split}\]

Since this last statement is clearly true for \(n\geq 1\), the initial statement \(b_{n+1}\leq b_n\) must also be true for \(n\geq 1\).

Since \(b_n=\tfrac{1}{n}\) satisfies both necessary conditions (the limit and the inquality), we are able to conclude that the original alternating series converges by the Alternating Series Test.

(Make sure you state which test you are using.)

Example 2#

Determine if the following series is convergent or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}2n^2}{5n^2-3} \]

(Click to see the steps.)

We see the alternating term \((-1)^{n+1}\) in our series, so we suspect the Alternating Series Test might be applicable. We begin by separating off this alternating piece, and writing the series as:

\[ \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}2n^2}{5n^2-3} = \sum_{n=1}^{\infty} (-1)^{n+1}\cdot \dfrac{2n^2}{5n^2-3} \]

This helps us identify the \(b_n\) term:

\[ b_n = \dfrac{2n^2}{5n^2-3} \]

The first condition we will check is that the limit of \(b_n\) is \(0\).

\[ \lim_{n\to\infty}b_n = \lim_{n\to\infty} \dfrac{2n^2}{5n^2-3} =\lim_{n\to\infty} \dfrac{\tfrac{2n^2}{n^2}}{\tfrac{5n^2}{n^2}-\tfrac{3}{n^2}} = \lim_{n\to\infty} \dfrac{2}{5-\tfrac{3}{n^2}} = \dfrac{2}{5-0} = \dfrac{2}{5} \]

Using our usual technique of dividing everything by the largest power of \(n\) in the denominator.

However, we see that:

\[ \lim_{n\to\infty}b_n = \dfrac{2}{5} \neq 0 \]

and so the limit condition of the alternating series test is not satisfied.

Since \(b_n= \tfrac{2n^2}{5n^2-3}\) does not satisfy the necessary limit condition, the alternating series test cannot be used.

However, we can instead use the Test for Divergence, since:

\[ \lim_{n\to\infty}b_n = \lim_{n\to\infty}\dfrac{2n^2}{5n^2-3} = \dfrac{2}{5} \neq 0 \]

means:

\[ \lim_{n\to\infty}a_n = \lim_{n\to\infty} \left((-1)^{n+1}\cdot \dfrac{2n^2}{5n^2-3}\right) \quad \text{ DNE} \]

Since this limit does not exist, the Test for Divergence tells us the original alternating seires diverges.

Non-Zero Limit#

What happens if we have a series where the \(b_n\) terms do not satisfy one or both of the conditions? Then the Alternating Series Test cannot be used.

This means that if you show either:

\[ b_{n+1}\geq b_n \qquad \text{for all } n \]
\[ \lim_{b\to \infty} b_n \neq 0 \]

the Alternating Series Test just cannot be applied, since the necessary conditions are not met.

However, we can instead use the Test for Divergence, since:

\[ \lim_{n\to \infty} b_n\neq 0 \quad \longrightarrow \quad \lim_{n\to \infty} (-1)^{n-1}b_n \; \; \text{DNE} \]

Which means that the alternating series \(\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} b_n\) diverges by the Test for Divergence.

Example 3#

Determine if the following series is convergent or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}n}{n^2+3} \]

(Click to see the steps.)

We see the alternating term \((-1)^{n-1}\) in our series, so we suspect the Alternating Series Test might be applicable. We begin by separating off this alternating piece, and writing the series as:

\[ \sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}n}{n^2+3} = \sum_{n=1}^{\infty} (-1)^{n-1}\cdot \dfrac{n}{n^2+3} \]

This helps us identify the \(b_n\) term:

\[ b_n = \dfrac{n}{n^2+3} \]

The first condition we will show is that the limit of \(b_n\) is \(0\).

\[ \lim_{n\to\infty}b_n = \lim_{n\to\infty} \dfrac{n}{n^2+3} =\lim_{n\to\infty} \dfrac{\tfrac{n}{n^2}}{\tfrac{n^2}{n^2}+\tfrac{3}{n^2}} = \lim_{n\to\infty} \dfrac{\tfrac{1}{n}}{\tfrac{1}{1}+\tfrac{3}{n^2}} = \dfrac{0}{1+0} = 0 \]

Using our usual technique of dividing everything by the largest power of \(n\) in the denominator.

Next, we show that the sequence \(\{b_n\}\) is decreasing, or more specifically that \(b_{n+1}\leq b_n\) eventually for all \(n\). To do this we will start by writing out the individual \(b_n\) and \(b_{n+1}\) terms:

\[ b_n = \dfrac{n}{n^2+3} \qquad \text{and}\qquad b_{n+1}= \dfrac{(n+1)}{(n+1)^2+3} \]

To actually prove the inequality we start with what we want and simplify until we get a statement that is clearly true:

\[\begin{split} & \qquad \quad b_{n+1} &\leq b_n \\[10pt] \iff & \dfrac{(n+1)}{(n+1)^2+3} & \leq \dfrac{n}{n^2+3} \\[10pt] \iff & (n+1)(n^2+3) & \leq n((n+1)^2+3) \\[10pt] \iff & n^3+n^2+3n+3 & \leq n(n^2+2n+1+3) \\[10pt] \iff & n^3+n^2+3n+3 & \leq n^3+2n^2+4n \\[10pt] \iff & \qquad \quad 3 & \leq n^2+n \\[10pt] \end{split}\]

Since this last statement is clearly true for \(n\geq 2\), the initial statement \(b_{n+1}\leq b_n\) must also be true for \(n\geq 2\).

Since \(b_n= \tfrac{n}{n^2+3}\) satisfies both necessary conditions (the limit and the inquality), we are able to conclude that the original alternating series converges by the Alternating Series Test.

(Make sure you state which test you are using.)