Successive terms in an alternating series have opposite sign. Under what conditions will an alternating series converge? Which of these series will converge? Why?

1. \(1-1+1-1+1-1+1-\cdots\)

2. \(-1 + 2 - 3 + 4 - 5 + 6 - \cdots\)

3. \(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots\)

4. \(1 - \frac{1}{2} + 3 - \frac{1}{4} + 5 - \frac{1}{6} + \cdots\)

Series: \(\displaystyle 5 \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} = 5 - \frac{5}{2} + \frac{5}{3} - \frac{5}{4} + \frac{5}{5} - \cdots\)

Seq. of partial sums: \(5, 2.5, 4.1666, 2.9166, 3.9166, \ldots\)

Theorem: Let \(\lbrace a_n \rbrace\) be a positive, decreasing sequence with \(\displaystyle \lim_{n \to \infty} a_n = 0\). Then \(\displaystyle \sum_{n=1}^{\infty} (-1)^n a_n\) and \(\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1} a_n\) both converge.

Explain why each of the validity conditions is necessary.

Does the alternating series test apply? What conclusions can be drawn?

1. \(\displaystyle \sum_{n=1}^{\infty} (-1)^n \frac{\ln(n)}{n}\)

2. \(\displaystyle \sum_{n=1}^{\infty} \frac{\sin(n)}{n}\)

3. \(\displaystyle \sum_{n=1}^{\infty} \frac{\cos(\pi n)}{n}\)

4. \(\displaystyle \sum_{n=1}^{\infty} (-1)^n \frac{n+1}{2n}\)

Series: \(\displaystyle \sum_{n=1}^{\infty} (-1)^n \frac{n+1}{2n} = -1 + \frac{3}{4} - \frac{2}{3} + \frac{5}{8} - \cdots\)

Partial sums: -1, -0.25, -0.9166, -0.2916, -0.8916,…

If possible, write formulas for the series and the sequence of partial sums, graph the sequence of partial sums, find what the series converges to or find a numerical estimate accurate to \(0.001\) using \(s_n\) and \(a_{n+1}\).

1. \(1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \cdots\)

2. \(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots\)

• Tip: Use a spreadsheet to create the sequence (\(a_n\)), sequence of partial sums (\(s_n\)), and a chart that shows the sequence of partial sums.
• Once you know what it converges to (\(L\)), use the spreadsheet to compute the errors.

The limit \(L\) is between \(s_n\) and \(s_{n+1}\).

Thus, \(|s_n - L| < a_{n+1}\) since \(s_n\) is within distance \(a_{n+1}\) of the limit \(L\).

Definition: A series \(\sum a_n\) converges absolutely if \(\sum |a_n|\) converges. A series \(\sum a_n\) converges conditionally if \(\sum a_n\) converges but \(\sum |a_n|\) diverges (i.e., it is convergent, but not absolutely convergent).

Examples

1. \(\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}\) converges absolutely.

2. \(\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\) converges conditionally.

3. If a series \(\sum a_n\) converges absolutely, then \(\sum a_n\) ( must / may / cannot ) converge.

Theorem: Suppose \(\displaystyle \sum_{n=1}^\infty |a_n|\) converges. Then

1. \(\displaystyle \sum_{n=1}^\infty a_n\) converges, and

2. if \(\displaystyle \lbrace b_n \rbrace\) is any rearrangement of \(\lbrace a_n \rbrace\), then \[\sum_{n=1}^\infty b_n = \sum_{n=1}^\infty a_n.\]

Determine whether each series converges, converges conditionally, converges absolutely, or diverges. Be sure to state what convergence tests you're using and check validity conditions.

1. \(\displaystyle \sum_{n=1}^{\infty} \frac{3n+5}{n^2-3n+1}\)

2. \(\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}\)

3. \(\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1}\frac{1}{\sqrt{n}}\)

Does a partial sum (which has finitely many terms!) change value if its terms are grouped or ordered differently? What about infinite sums – does their value change if the terms are rearranged?

Alice in Wonderland (by Lewis Carroll)

"Alice has fallen down the rabbit hole and eaten a cake that has shrunk her to a height of just 3 inches. The Caterpillar enters, smoking a hookah pipe, and shows Alice a mushroom that can restore her to her proper size. But one side of the mushroom stretches her neck, while another shrinks her torso, so she must eat exactly the right balance to regain her proper size and proportions." (Summary by Keith Devlin)

Rearranging finitely many terms of an infinite series does not change the sum of the series because there will be some partial sum \(s_n\) that contains all of the rearranged terms, \(s_n\) is a finite sum unaffected by rearrangement, all the partial sums \(s_k\) for \(k > n\) are thus unaffected by the rearrangement, and thus \(\displaystyle\sum a_n = \lim_{n\to\infty} s_n\) is unchanged by rearranging finitely many terms.

Rearranging infinitely many terms of a conditionally convergent series can change the sum of the series! When you rearrange infinitely many terms, you can change the value of infinitely many partial sums (and hence the sum of the series).

Rearranging infinitely many terms of an absolutely convergent series does not change the sum of the series!

\[\sum_{n=1}^\infty \frac{(-1)^n}{n} \ne \sum_{k=1}^\infty \frac{1}{2k} - \sum_{k=1}^\infty \frac{1}{2k-1}\]

The alternating series (left) converges, while the "even term" and "odd term" series both diverge to infinity. (Remember: \(\infty - \infty\) is an indeterminate form!). Notice that we have rearranged infinitely many terms when going from the LHS to the RHS.

Consider \(\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{n} = \lim_{n\to\infty} \left(-1+\frac{1}{2}-\frac{1}{3}+\cdots\pm\frac{1}{n}\right)\). Convince yourself it converges to a negative number.

Now rearrange the terms:

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\) (greater than 1)
\(-1\) (positive partial sum)
\(\frac{1}{10}+\frac{1}{12}+\cdots+\frac{1}{18}\) (greater than 1/3)
\(-\frac{1}{3}\) (positive partial sum)
\(\frac{1}{20}+\frac{1}{22}+\cdots+\frac{1}{28}\) (greater than 1/5)
\(-\frac{1}{5}\) (positive partial sum)
…all partial sums are positive, so the rearranged series either converges to a positive number or diverges!