Lecture 31 Power Series

Text References: Course notes pp. 137-139 & Rogawski 10.6

31.1 Recap

Last time, we learned about Ratio Test and the Root Test.

Exercise 31.1 Determine whether the series \(\displaystyle \sum_{k=1}^\infty \frac{3^k}{k!}\) converges.

Solution. Using the Ratio test, we have

\[\begin{align*} \lim_{k\to\infty}\left | \frac{a_{k+1}}{a_k}\right| &= \lim_{k\to\infty}\left | \frac{\frac{3^{k+1}}{(k+1)!}}{\frac{3^k}{k!}}\right | \\ &= \lim_{k\to\infty} \frac{3}{k+1} = 0 < 1 \end{align*}\]

Therefore, by the Ratio Test, the series is absolutely convergent

31.2 Learning Objectives

  • Determine the radius of convergence of a power series.

31.3 Power Series

Let’s cast our memories back to Taylor series for a moment. We defined the Taylor series of a function \(f(x)\) centred at \(x_0\) to be the series \(\displaystyle \sum_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n\). When we tried to determine whether a function was in fact equal to its Taylor series, we can into some problems which, incidentally, started our discussion about series and convergence. We saw an example in Lecture 25 where the Taylor polynomial did better and better at approximating \(f(x)=\sin(x)\) at \(x=2\) and worse and worse at approximating a \(g(x)=\frac{1}{1+x}\) at the same point.

Now, things weren’t all bad in the example; the Taylor polynomial was doing well at approximating \(g(x)=\frac{1}{1+x}\) for certain values of \(x\). Put another way, the Taylor polynomial seemed to converge towards \(g(x)\) for some values of \(x\) and diverge from \(g(x)\) over a different set of values of \(x\). If this language reminds you of all the work we’ve done with series, you’re absolutely right!

The results that we’re going to develop are typically stated in terms of power series:

Definition 31.1 A power series centred at \(x_0\) is any series of the form \[\sum_{k=0}^\infty c_k(x-x_0)^k=c_0+c_1(x-x_0)+c_2(x-x_0)^2+\cdots\]

31.4 Convergence of Power Series

As we were discussing before, we’re interested in knowing for which values a power series converges. To do so, we’ll apply the Ratio Test:

\[\begin{align*} \lim_{k\to\infty}\left | \frac{a_{k+1}}{a_k}\right | &= \lim_{k\to\infty}\left | \frac{c_{k+1}(c-c_0)^{k+1}}{c_k(x-x_0)^k}\right| \\ &= \lim_{k\to\infty}\left |\frac{c_{k+1}}{c_k} \right | |x-x_0|\\ &= |x-x_0| \lim_{k\to\infty}\left |\frac{c_{k+1}}{c_k} \right | \end{align*}\]

The power series will converge absolutely if \[|x-x_0| \lim_{k\to\infty}\left |\frac{c_{k+1}}{c_k} \right |< 1\] that is, if \[|x-x_0| < \lim_{k\to\infty}\left |\frac{c_{k+1}}{c_k} \right |\]

Let \(\displaystyle R=\lim_{k\to\infty}\left |\frac{c_{k+1}}{c_k} \right |\). We call \(R\) the radius of convergence. We have the following result:

Theorem 31.1 Consider a power series \(\displaystyle \sum_{k=0}^\infty c_k(x-x_0)^k\) with radius of convergence \(R\).

  • If \(R=\infty\), then the series converges absolutely for all \(x\)
  • If \(R=0\), then the series converges only at \(x=x_0\).
  • If \(0<R<\infty\), then the series converges absolutely for \(x\in (x_0-R, x_0+R)\) and diverges for \(x<x_0-R\) and for \(x>x_0+R\). The test gives no conclusion at the points \(x=x_0\pm R\).

From the above theorem, we gather that there are two steps to determine the interval of convergence of a power series:

  1. Find the radius of convergence \(R\). This is typically done using the Ratio Test.
  2. If \(0<R<\infty\), verify the convergence at the endpoints.

Exercise 31.2 Find the values of \(x\) for which \(\displaystyle \sum_{k=0}^\infty \frac{x^k}{2^k}\) converges.

Solution. First, we apply the Ratio Test with \(a_k= \dfrac{x^k}{2^k}\) to find \(R\):

\[\begin{align*} \lim_{k\to\infty} \left | \frac{a_{k+1}}{a_k}\right | &= \lim_{k\to\infty}\left | \frac{x^{k+1}}{2^{k+1}} \right|\left | \frac{2^k}{x^k} \right| \\ &= \lim_{k\to\infty} \frac{1}{2}|x| = \frac{1}{2}|x| \end{align*}\]

We have \(R=\frac{1}{2}|x|\) and so \(R<1\) if \(|x|<2\); that is, the radius of convergence is \(R=2\).

Second, we need to check the endpoints. We have

  • \(\displaystyle \sum_{k=0}^\infty \frac{2^k}{2^k}=\sum_{k=0}^\infty 1\) which diverges; and
  • \(\displaystyle \sum_{k=0}^\infty \frac{(-2)^k}{2^k}=\sum_{k=0}^\infty (-1)^k\) which also diverges.

Therefore, the power series converges for \(|x|<2\), i.e. on the interval \((-2,2)\).

Exercise 31.3 Find the values of \(x\) for which \(\displaystyle \sum_{k=0}^\infty k!(x-2)^k\) converges.

Solution. First, we apply the Ratio Test with \(a_k= \dfrac{x^k}{2^k}\) to find \(R\):

\[\begin{align*} \lim_{k\to\infty} \left | \frac{a_{k+1}}{a_k}\right | &= \lim{k\to\infty}\left | \frac{(k+1)!(x-2)^{k+1}}{k!(x-2)^k} \right | \\ &= \lim_{k\to\infty} (k+1)|x-2|=\infty \end{align*}\]

We have \(R=\infty\) and so the power series converges for all values of \(x\).