*Definition.*An*alternating series*is a series whose terms are alternately positive and negative.- Examples: \[\begin{gather*} 1-\frac12+\frac13-\frac14+\frac15-\frac16+\dotsb=\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n}\\ -\frac12+\frac23-\frac34+\frac45-\frac56+\frac67-\dotsb=\sum_{n=1}^\infty(-1)^n\frac{n}{n+1}. \end{gather*}\]
- In general, \(\sum a_n\) is an alternating series, if for every \(n\), we have \[ a_n=(-1)^{n-1}b_n\text{ or }a_n=(-1)^nb_n \] where \(b_n>0\) for all \(n\). Note that we have \(b_n=|a_n|\).

*Alternating series test.*- Let \(\sum a_n\) be an alternating series, and let \(b_n=|a_n|\).
- Suppose that the sequence \(\{b_n\}\) is decreasing.
- Suppose also that we have \(\lim_{n\to\infty}b_n=0\).
- Then the series \(\sum a_n\) is convergent.

- Example. Consider the series \(\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n}\).
- Since \(\frac{1}{n}\ge\frac{1}{n+1}\) for all \(n\ge1\), the sequence \(\{b_n\}=\left\{\frac{1}{n}\right\}\) is decreasing.
- We have \(\lim_{n\to\infty}\frac{1}{n}=0\).
- Therefore, the series \(\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n}\) is convergent.

- Exercises. 11.5: 3, 5, 11, 13, 17, 19

*Alternating series estimation theorem.*Suppose that \(s=\sum(-1)^nb_n\), where \(\{b_n\}\) is a sequence that is positive, decreasing, and converges to 0.- Then for any \(n\), we have \[ |R_n|=|s-s_n|\le b_{n+1}. \]

- Example. Consider the series \(\sum_{n=0}^\infty(-1)^n\frac{2^n}{3\cdot 5^n}\).
- You can check that is satisfies the conditions of the Alternating series test.
- Let's find an approximation of \(s=\sum_{n=0}^\infty(-1)^n\frac{2^n}{3\cdot 5^n}\) correct to 3 decimal places.
- For this, we need to find an \(n\) such that \(\frac{2^n}{3\cdot 5^n}<0.0005\).
- For \(n=8\), we have \(\frac{2^8}{3\cdot 5^8}\approx0.0002\).
- This means that \(s_8\approx0.5554\) is correct up to 3 decimal places.
- Click here for the minimalistic SAGE code I used to perform the approximation.

- Exercises. 11.5: 23, 27.

*Definition.*Let \(\sum a_n\) be a series.- Suppose that the series \(\sum|a_n|\) is convergent.
- Then we say that the series \(\sum a_n\) is
*absolutely convergent*.

- Examples.
- The series \[ \sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n^3}=1-\frac18+\frac{1}{27}-\frac{1}{64}+\dotsb \] is absolutely convergent, because the series \[ \sum_{n=1}^\infty\frac{1}{n^3} \] is convergent as \(3>1\).
- We have seen that by the Alternating series test, the alternating harmonic series \(\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n}\) is convergent. But it's not absolutely convergent, because the harmonic series \(\sum_{n=1}^\infty\frac{1}{n}\) is divergent.

*Definition.*If a series is convergent but not absolutely convergent, we say it is*conditionally convergent*.*Theorem.*If a series is absolutely convergent, then it is convergent.- Example. Consider the series \(\sum_{n=1}^\infty\frac{\sin n}{n^2}\).
- Since we have \[ 0\le\frac{|\sin n|}{n^2}\le\frac{1}{n^2}, \] the Comparison theorem implies that \(\sum_{n=1}^\infty\frac{\sin n}{n^2}\) is absolutely convergent, and therefore it is convergent.

*Ratio test.*Let \(\sum a_n\) be a series.- If \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L<1\), then \(\sum a_n\) is absolutely convergent.
- If \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L>1\) or \(\infty\), then \(\sum a_n\) is divergent.
- If \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1\) or divergent, then the Ratio test is inconclusive.

- Example. Consider the series \(\sum_{n=1}^\infty\frac{n^2}{3^n}\).
- We have \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)^2}{n^2}\cdot\frac{3^n}{3^{n+1}}\to\frac13<1\), therefore the series is convergent.

- Example. Let's revisit the series \(\sum_{n=1}^\infty\frac{n^n}{n!}\).
- We have \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}\)==()
^{n=(1+)}ne>1$, therefore the series is divergent.

- We have \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}\)==()
- Exercises. 11.6: 7, 9, 17, 21.

*Root test.*Let \(\sum a_n\) be a series.- If \(\lim_{n\to\infty}\sqrt[n]{|a_n|}=L<1\), then \(\sum a_n\) is absolutely convergent.
- If \(\lim_{n\to\infty}\sqrt[n]{|a_n|}=L>1\) or \(\infty\), then \(\sum a_n\) is divergent.
- If \(\lim_{n\to\infty}\sqrt[n]{|a_n|}=1\) or divergent, then the Root test is inconclusive.
- Note that the Ratio and Root tests yield a limit of 1 at the same time. That is, if the limit is 1 for one of the tests, then you shouldn't try the other test, but some other method.

- Example. Consider the series \(\sum_{n=1}^\infty\left(\frac{n-1}{n^2+2n+8}\right)^{2n}\).
- Since we have \(\sqrt[n]{a_n}=\left(\frac{n-1}{n^2+2n+8}\right)^2\to0\), the series is absolutely convergent.

- Exercises. 11.6: 27, 29.
- More exercises. 11.6: 31, 33, 35, 37.

- Now let's look at something very weird.
- Let \(\sum a_n\) be a series. A
*rearrangement*of \(\sum a_n\) is a series \(\sum b_n\), with exactly the same terms, but in a different order. For example, we could have \[ \sum b_n=a_2+a_4+a_1+a_5+a_3+\dotsb \]*Theorem.*If \(\sum a_n\) is absolutely convergent, then any rearrangement of it will have the same sum.

- It turns out that this is not true if \(\sum a_n\) is conditionally convergent!
*Theorem.*Let \(\sum a_n\) be a conditionally convergent series, and \(r\) any real number. Then there exists a rearrangement of \(\sum a_n\) with sum \(r\).- If you're interested, you can find a proof sketch in 11.6.52.

- Example. Consider the alternating harmonic series \(\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n}\).
- We have \[\begin{equation}\tag{I} 1-\frac12+\frac13-\frac14+\frac15-\dotsb=\ln2. \end{equation}\]
- Multiplying by \(\frac12\), we get \[ \frac12-\frac14+\frac16-\frac18+\frac{1}{10}-\dotsb=\frac12\ln2. \]
- We can insert zeros in this to get \[\begin{equation}\tag{II} 0+\frac12+0-\frac14+0+\frac16+\dotsb=\frac12\ln2. \end{equation}\]
- Now (I)\(+\)(II) is \[ 1+\frac13-\frac12+\frac15+\frac17-\frac14+\dotsb=\frac32\ln2, \] which you can check is a rearrangement of \(\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n}\).