Class 24 - Absolute convergence test + Alternating series test (10/31/2017, Tuesday)
Logistics
Exam 2- 11/13 Mon
WebHW
- 9.3 due 11/1 Wed
- 9.4 due 11/6 Mon
TeamHW
- Due today
Lecture
We start by an example:
Example. Converge/diverge?
\[\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{2}}.\]
This is somewhat hard to see right away. The key is to note that the series is convergent if and only if its "tail gets small". That is, we want to show that
\[\lim_{N \rightarrow \infty}\left|\sum_{n=N}^{\infty}\frac{(-1)^{n-1}}{n^{2}}\right| = 0.\]
Now, this is not hard to show because
\[\lim_{N \rightarrow \infty}\left|\sum_{n=N}^{\infty}\frac{(-1)^{n-1}}{n^{2}}\right| \leq \lim_{N \rightarrow \infty}\sum_{n=N}^{\infty}\frac{1}{n^{2}} = 0.\]
Using the $p$-series with $p = 2 > 1$. This generalizes to the following:
Absolute convergence implies conditional convergence. If $\sum_{n=1}^{\infty}|a_{n}|$ converges so does $\sum_{n=1}^{\infty}a_{n}$.
Warning. The above test does not tell you about the divergence $\sum_{n=1}^{\infty}a_{n}$. For example take $a_{n} = (-1)^{n-1}/n$. The absolute series $\sum_{n=1}^{\infty}|a_{n}| = \sum_{n=1}^{\infty}1/n$ diverges as it is a $p$-series with $p = 1 \leq 1$, but it turns out that the following series converges:
\[\sum_{n=1}^{\infty}a_{n} = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}.\]
We follow an explanation given in the Wikipedia page. To see this, we again need to consider the partial sums in a finer manner:
\[s_{m} = \sum_{n=1}^{m}\frac{(-1)^{n-1}}{n} = 1 - \frac{1}{2} + \cdots + (-1)^{m-l}\frac{1}{m}.\]
The odd terms $s_{2m-1}$ is a monotone decreasing sequence:
\[1, (1 - 1/2) + 1/3, (1 - 1/2) + (1/3 -1/4) + 1/5, \dots.\]
The sequence is bounded below by $0$, so it must converge. Similarly the even partial sums $s_{2m}$ must converge. Moreover, note that
\[|s_{2m} - s_{2m-1}| = 1/2m,\]
which goes to $0$ as $m \rightarrow \infty$. This implies that $s_{2m}$ and $s_{2m-1}$ converges to the same limit, so $s_{m}$ must converge to that limit, which shows that our infinite sum $\sum_{n=1}^{\infty}(-1)^{n-1}/n$ is convergent. Essentially the same argument shows:
Alternating series test. If
Exercise*. Do #3 of Exam 3, Fall 2016.
Exercise*. Do #9 d of Exam 3, Fall 2016.
Exercise*. Do #2 a of Exam 3, Winter 2016.
Exercise*. Do #5 of Exam 3, Winter 2016.
Exercise* (Introduction to ratio test). Do #11 of Exam 3, Winter 2016.
Example. Converge/diverge?
\[\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{2}}.\]
This is somewhat hard to see right away. The key is to note that the series is convergent if and only if its "tail gets small". That is, we want to show that
\[\lim_{N \rightarrow \infty}\left|\sum_{n=N}^{\infty}\frac{(-1)^{n-1}}{n^{2}}\right| = 0.\]
Now, this is not hard to show because
\[\lim_{N \rightarrow \infty}\left|\sum_{n=N}^{\infty}\frac{(-1)^{n-1}}{n^{2}}\right| \leq \lim_{N \rightarrow \infty}\sum_{n=N}^{\infty}\frac{1}{n^{2}} = 0.\]
Using the $p$-series with $p = 2 > 1$. This generalizes to the following:
Absolute convergence implies conditional convergence. If $\sum_{n=1}^{\infty}|a_{n}|$ converges so does $\sum_{n=1}^{\infty}a_{n}$.
Warning. The above test does not tell you about the divergence $\sum_{n=1}^{\infty}a_{n}$. For example take $a_{n} = (-1)^{n-1}/n$. The absolute series $\sum_{n=1}^{\infty}|a_{n}| = \sum_{n=1}^{\infty}1/n$ diverges as it is a $p$-series with $p = 1 \leq 1$, but it turns out that the following series converges:
\[\sum_{n=1}^{\infty}a_{n} = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}.\]
We follow an explanation given in the Wikipedia page. To see this, we again need to consider the partial sums in a finer manner:
\[s_{m} = \sum_{n=1}^{m}\frac{(-1)^{n-1}}{n} = 1 - \frac{1}{2} + \cdots + (-1)^{m-l}\frac{1}{m}.\]
The odd terms $s_{2m-1}$ is a monotone decreasing sequence:
\[1, (1 - 1/2) + 1/3, (1 - 1/2) + (1/3 -1/4) + 1/5, \dots.\]
The sequence is bounded below by $0$, so it must converge. Similarly the even partial sums $s_{2m}$ must converge. Moreover, note that
\[|s_{2m} - s_{2m-1}| = 1/2m,\]
which goes to $0$ as $m \rightarrow \infty$. This implies that $s_{2m}$ and $s_{2m-1}$ converges to the same limit, so $s_{m}$ must converge to that limit, which shows that our infinite sum $\sum_{n=1}^{\infty}(-1)^{n-1}/n$ is convergent. Essentially the same argument shows:
Alternating series test. If
- $a_{n}$ alternates the sign,
- $|a_{n}|$ is decreasing, and
- $\lim_{n\rightarrow \infty}a_{n} = 0$,
Exercise*. Do #3 of Exam 3, Fall 2016.
Exercise*. Do #9 d of Exam 3, Fall 2016.
Exercise*. Do #2 a of Exam 3, Winter 2016.
Exercise*. Do #5 of Exam 3, Winter 2016.
Exercise* (Introduction to ratio test). Do #11 of Exam 3, Winter 2016.
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