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Section 9.6 Alternating Series; Conditional and Absolute Convergence

If we recall, the harmonic series test is:

\begin{equation*} \sum_{n=1}^{\infty} \dfrac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots \quad \text{diverges} \end{equation*}

If we make this the alternating harmonic series, we get a different result:

\begin{equation*} \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \quad \text{converges} \end{equation*}

If we try to prove this using the ratio test it fails and the result is inconclusive. Inorder to prove this we will have to use the alternating series test.

Subsection 9.6.1 Alternating Series Test

An alternating series of the form

\begin{equation*} \sum_{n=1}^{\infty} (-1)^{n+1} a_n = a_1 + a_2 + a_3 + a_4 + \dots \end{equation*}

converges if:

  • \(0 \lt a_{n+1} \lt a_n\) for all \(n\) (strictly decreasing)

  • \(\displaystyle \displaystyle \lim_{n\to\infty} a_n = 0\)

Example 9.6.1.

Let's apply this test to the alternating harmonic series:

\begin{equation*} a_n = \frac{1}{n}, \quad \frac{1}{n+1} \end{equation*}

  1. \(\displaystyle \dfrac{a_{n+1}}{a_n} = \dfrac{\frac{1}{n+1}}{\frac{1}{n}} = \dfrac{n}{n+1} \lt 1 \quad \text{strictly decreasing}\)

  2. \(\displaystyle \displaystyle \lim_{n\to\infty} a_n = \displaystyle \lim_{n\to\infty} \dfrac{1}{n} = 0\)

\begin{equation*} \boxed{ \sum \dfrac{(-1)^{n+1}}{n} \text{ converges by the alternating series test} } \end{equation*}

Subsection 9.6.2 Absolute and Conditional Convergence

We say \(\displaystyle \sum a_n\) is

  • absolutely convergent if \(\displaystyle \mid \sum a_n \mid \) and \(\displaystyle \sum a_n\) both converge

  • conditionally convergent if \(\displaystyle \mid \sum a_n \mid \) diverges but \(\displaystyle \sum a_n\) converges

    RECALL: \(a_n \leq \mid a_n \mid\)

  • additionally, if \(\displaystyle \mid \sum a_n \mid \) converges, \(\displaystyle \sum a_n\) converges by the comparison test

Example 9.6.2.

Determine if the following are absolutely or conditionally convergent:

  1. \begin{equation*} \sum \dfrac{(-1)^{n+1}}{n^2} \end{equation*}
    \begin{align*} \sum \bigg| \dfrac{(-1)^{n+1}}{n^2} \bigg| \amp = \sum \dfrac{1}{n^2} \\ \text{converges} \amp \text{ p-series where } p = 2 \end{align*}
    \begin{equation*} \boxed{\text{absolutely convergent}} \end{equation*}
  2. \begin{equation*} \sum \dfrac{(-1)^{n+1}}{n} \end{equation*}
    \begin{align*} \sum \bigg| \dfrac{(-1)^{n+1}}{n} \bigg| \amp = \sum \dfrac{1}{n} \\ \text{ diverges} \amp \text{ harmonic series}\\ \amp \\ \sum \dfrac{(-1)^{n+1}}{n} \amp \\ \text{converges} \amp \text{ alternating harmonic series} \end{align*}
    \begin{equation*} \boxed{\text{ conditionally convergent}} \end{equation*}