Learning Outcomes
- Determine the interval of convergence for a given power series.
Section 9.2 Convergence of Power Series (PS2)
Subsection 9.2.1 Activities
Activity 9.2.1.
Consider the series \(\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n\) where \(x\) is a real number.
(a)
If \(x=2\text{,}\) then \(\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{2^n}{n!}\text{.}\) What can be said about this series?
- The techniques we have learned so far allow us to conclude that \(\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{2^n}{n!}\) converges.
- The techniques we have learned so far allow us to conclude that \(\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{2^n}{n!}\) diverges.
- None of the techniques we have learned so far allow us to conclude whether \(\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{2^n}{n!}\) converges or diverges.
(b)
If \(x=-100\text{,}\) then \(\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{(-100)^n}{n!}\text{.}\) What can be said about this series?
- The techniques we have learned so far allow us to conclude that \(\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{(-100)^n}{n!}\) converges.
- The techniques we have learned so far allow us to conclude that \(\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{(-100)^n}{n!}\) diverges.
- None of the techniques we have learned so far allow us to conclude whether \(\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{(-100)^n}{n!}\) converges or diverges.
(c)
Suppose that \(x\) were some arbitrary real number. What can be said about this series?
- The techniques we have learned so far allow us to conclude that \(\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n\) converges.
- The techniques we have learned so far allow us to conclude that \(\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n\) diverges.
- None of the techniques we have learned so far allow us to conclude whether \(\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n\) converges or diverges.
Remark 9.2.2.
Consider a power series \(\displaystyle\sum c_n(x-a)^n\text{.}\) Recall from Fact 8.7.6 that if
\begin{align*}
\displaystyle \lim_{n\to \infty} \left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\right| & < 1
\end{align*}
then \(\displaystyle\sum c_n(x-a)^n\) converges.
Then recall:
\begin{align*}
\displaystyle \lim_{n\to \infty} \left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\right| & = \displaystyle \lim_{n\to \infty} \left|\frac{c_{n+1}(x-a)}{c_n}\right|\\
&=\displaystyle \lim_{n\to \infty} |x-a|\left|\frac{c_{n+1}}{c_n}\right| \\
&=\displaystyle |x-a|\lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right|.
\end{align*}
Activity 9.2.3.
Consider \(\displaystyle\sum_{n=0}^\infty \frac{1}{n^2+1}x^n.\)
(a)
Letting \(c_n=\frac{1}{n^2+1}\text{,}\) find \(\displaystyle \lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right|\text{.}\)
(b)
For what values of \(x\) is \(\displaystyle |x|\lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right| < 1\text{?}\)
- \(x < 1\text{.}\)
- \(0\leq x < 1\text{.}\)
- \(-1 < x < 1\text{.}\)
(c)
If \(x=1\text{,}\) does \(\displaystyle\sum_{n=0}^\infty \frac{1}{n^2+1}x^n\) converge?
(d)
If \(x=-1\text{,}\) does \(\displaystyle\sum_{n=0}^\infty \frac{1}{n^2+1}x^n\) converge?
(e)
Which of the following describe the values of \(x\) for which \(\displaystyle\sum_{n=0}^\infty \frac{1}{n^2+1}x^n\) converges?
- \((-1,1)\text{.}\)
- \([-1,1)\text{.}\)
- \((-1,1]\text{.}\)
- \([-1,1]\text{.}\)
Activity 9.2.4.
Consider \(\displaystyle\sum_{n=0}^\infty \frac{2^n}{5^n}(x-2)^n.\)
(a)
Letting \(c_n=\frac{2^n}{5^n}\text{,}\) find \(\displaystyle \lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right|\text{.}\)
(b)
For what values of \(x\) is \(\displaystyle |x-2|\lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right| < 1\text{?}\)
- \(-\frac{2}{5} < x < \frac{2}{5}\text{.}\)
- \(\frac{8}{5} < x < \frac{12}{5}\text{.}\)
- \(-\frac{5}{2} < x < \frac{5}{2}\text{.}\)
- \(-\frac{1}{2} < x < \frac{9}{2}\text{.}\)
(c)
If \(x=\frac{9}{2}\text{,}\) does \(\displaystyle\sum_{n=0}^\infty \frac{2^n}{5^n}(x-2)^n\) converge?
(d)
If \(x=-\frac{1}{2}\text{,}\) does \(\displaystyle\sum_{n=0}^\infty \frac{2^n}{5^n}(x-2)^n\) converge?
(e)
Which of the following describe the values of \(x\) for which \(\displaystyle\sum_{n=0}^\infty \frac{2^n}{5^n}(x-2)^n\) converges?
- \((-\frac{1}{2},\frac{9}{2})\text{.}\)
- \([-\frac{1}{2},\frac{9}{2})\text{.}\)
- \((-\frac{1}{2},\frac{9}{2}]\text{.}\)
- \([-\frac{1}{2},\frac{9}{2}]\text{.}\)
Activity 9.2.5.
Consider \(\displaystyle\sum_{n=0}^\infty \frac{n^2}{n!}\left(x+\frac{1}{2}\right)^n.\)
(a)
Letting \(c_n=\frac{n^2}{n!}\text{,}\) find \(\displaystyle \lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right|\text{.}\)
(b)
For what values of \(x\) is \(\displaystyle \left|x+\frac{1}{2}\right|\lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right| < 1\text{?}\)
- \(0\leq x < \infty\text{.}\)
- All real numbers.
(c)
What describes the values of \(x\) for which \(\displaystyle\sum_{n=0}^\infty \frac{n^2}{n!}\left(x+\frac{1}{2}\right)^n\) converges?
Fact 9.2.6.
Given the power series \(\displaystyle\sum c_n(x-a)^n\text{,}\) the center of convergence is \(x=a\text{.}\) The radius of convergence is
\begin{equation*}
r=\frac{1}{\displaystyle\lim_{n\to\infty} \left| \frac{c_{n+1}}{c_n} \right|}.
\end{equation*}
If \(\displaystyle\lim_{n\to\infty} \left| \frac{c_{n+1}}{c_n} \right|=0\text{,}\) we say that \(r=\infty\text{.}\)
The interval of convergence represents all possible values of \(x\) for which \(\displaystyle\sum c_n(x-a)^n\) converges, which is of the form:
- \(\displaystyle (a-r, a+r)\)
- \(\displaystyle [a-r, a+r)\)
- \(\displaystyle (a-r, a+r]\)
- \(\displaystyle [a-r, a+r]\)
Depending on if \(\displaystyle\sum c_n(x-a)^n\) converges when \(x=a-r\) or \(x=a+r\text{.}\)
If \(r=\infty\text{,}\) the interval of convergence is all real numbers.
Activity 9.2.7.
Find the center of convergence, radius of convergence, and interval of convergence for the series:
\begin{equation*}
\sum_{n=0}^\infty \frac{3^{n} \left(-1\right)^{n} {\left(x - 1\right)}^{n}}{n!}.
\end{equation*}
Activity 9.2.8.
Find the center of convergence, radius of convergence, and interval of convergence for the series:
\begin{equation*}
\sum_{n=0}^\infty \frac{3^{n} {\left(x + 2\right)}^{n}}{n}.
\end{equation*}
Activity 9.2.9.
Consider the power series \(\displaystyle \sum_{n=0}^\infty \frac{2^n+1}{n3^n}\left(x+1\right)^n\text{.}\)
(a)
What is the center of convergence for this power series?
(b)
What is the radius of convergence for this power series?
(c)
What is the interval of convergence for this power series?
(d)
If \(x=-0.5\text{,}\) does this series converge? (Use the interval of convergence.)
(e)
If \(x=1\text{,}\) does this series converge? (Use the interval of convergence.)
Subsection 9.2.2 Videos
Subsection 9.2.3 Exercises
Exercises available at
https://tbil.org/calculus/preview/exercises/#/bank/PS2/
