8.3: Radius of Convergence of a Power Series
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Skills to Develop
 Explain the radius of convergence of a power series
We’ve developed enough machinery to look at the convergence of power series. The fundamental result is the following theorem due to Abel.
Theorem \(\PageIndex{1}\)
Suppose \( \displaystyle \sum_{n=0}^{\infty } a_nc^n\) converges for some nonzero real number \(c\). Then \( \displaystyle \sum_{n=0}^{\infty } a_n x^n\) converges absolutely for all \(x\) such that \(x < c\).
 Sketch of Proof

To prove Theorem \(\PageIndex{1}\) ﬁrst note that by \(\lim_{n \to \infty } a_n c^n = 0\). Thus (\(a_nc^n\)) is a bounded sequence. Let \(B\) be a bound: \(ac^n≤ B\). Then
\[\left  a_n x^n \right  = \left  a_n c^n\cdot \left ( \frac{x}{c} \right )^n \right  \leq B \left  \frac{x}{n} \right ^c\]
We can now use the comparison test.
Exercise \(\PageIndex{1}\)
Prove Theorem \(\PageIndex{1}\).
Corollary \(\PageIndex{1}\)
Suppose \( \displaystyle \sum_{n=0}^{\infty } a_nc^n\) diverges for some real number \(c\). Then \( \displaystyle \sum_{n=0}^{\infty } a_nx^n\) diverges for all \(x\) such that \(x > c\).
Exercise \(\PageIndex{2}\)
Prove Corollary \(\PageIndex{1}\).
As a result of Theorem \(\PageIndex{1}\) and Corollary \(\PageIndex{1}\), we have the following: either \( \displaystyle \sum_{n=0}^{\infty } a_n x^n\) converges absolutely for all \(x\) or there exists some nonnegative real number \(r\) such that \( \displaystyle \sum_{n=0}^{\infty } a_n x^n\) converges absolutely when \(x < r\) and diverges when \(x > r\). In the latter case, we call \(r\) the radius of convergence of the power series \( \displaystyle \sum_{n=0}^{\infty } a_n x^n\). In the former case, we say that the radius of convergence of \( \displaystyle \sum_{n=0}^{\infty } a_n x^n\) is \(∞\). Though we can say that \( \displaystyle \sum_{n=0}^{\infty } a_n x^n\) converges absolutely when \(x < r\), we cannot say that the convergence is uniform. However, we can come close. We can show that the convergence is uniform for \(x≤ b < r\). To see this we will use the following result.
Theorem \(\PageIndex{2}\): The WeierstrassM Test
Let \(\left ( f_n \right )_{n=1}^{\infty }\) be a sequence of functions deﬁned on \(S ⊆ \mathbb{R}\) and suppose that \(\left ( M_n \right )_{n=1}^{\infty }\) is a sequence of nonnegative real numbers such that
\[f_n(x)≤ M_n, ∀x ∈ S, n = 1,2,3,....\]
If \( \displaystyle \sum_{n=1}^{\infty } M_n\) converges then \( \displaystyle \sum_{n=1}^{\infty } f_n(x)\) converges uniformly on \(S\) to some function (which we will denote by \(f(x)\)).
 Sketch of Proof

Since the crucial feature of the theorem is the function \(f(x)\) that our series converges to, our plan of attack is to ﬁrst deﬁne \(f(x)\) and then show that our series, \( \displaystyle \sum_{n=1}^{\infty } f_n(x)\), converges to it uniformly.
First observe that for any \(x ∈ S\), \( \displaystyle \sum_{n=1}^{\infty } f_n(x)\) converges by the Comparison Test (in fact it converges absolutely) to some number we will denote by \(f(x)\). This actually deﬁnes the function \(f(x)\) for all \(x ∈ S\). It follows that \( \displaystyle \sum_{n=1}^{\infty } f_n(x)\) converges pointwise to \(f(x)\).
Next, let \(ε > 0\) be given. Notice that since \( \displaystyle \sum_{n=1}^{\infty }M _n\) converges, say to \(M\), then there is a real number, \(N\), such that if \(n > N\), then
\[\sum_{k=n+1}^{\infty } M_k = \left  \sum_{k=n+1}^{\infty } M_k \right  = \left  M  \sum_{k=1}^{n} M_k \right  < \varepsilon\]
You should be able to use this to show that if \(n > N\), then
\[\left  f(x)  \sum_{k=1}^{n} f_k(x) \right  < \varepsilon ,\; \forall x\; \epsilon \; S\]
Exercise \(\PageIndex{3}\)
Use the ideas above to provide a formal proof of Theorem \(\PageIndex{2}\).
Exercise \(\PageIndex{4}\)
 Show that the Fourier series \[\sum_{k=0}^{\infty } \frac{(1)^k}{(2k+1)^2} \sin ((2k+1)\pi x)\] converges uniformly on \(\mathbb{R}\).
 Does its diﬀerentiated series converge uniformly on \(\mathbb{R}\)? Explain.
Exercise \(\PageIndex{5}\)
Observe that for all \(x ∈ [1,1] x ≤ 1\). Identify which of the following series converges pointwise and which converges uniformly on the interval \([1,1]\). In every case identify the limit function.
 \( \displaystyle \sum_{n=1}^{\infty }\left ( x^n  x^{n1} \right )\)
 \( \displaystyle \sum_{n=1}^{\infty }\frac{\left ( x^n  x^{n1} \right )}{n}\)
 \( \displaystyle \sum_{n=1}^{\infty }\frac{\left ( x^n  x^{n1} \right )}{n^2}\)
Using the WeierstrassM test, we can prove the following result.
Theorem \(\PageIndex{3}\)
Suppose \( \displaystyle \sum_{n=0}^{\infty }a_nx^n\) has radius of convergence \(r\) (where \(r\) could be \(∞\) as well). Let \(b\) be any nonnegative real number with \(b < r\). Then \( \displaystyle \sum_{n=0}^{\infty }a_nx^n\) converges uniformly on \([b,b]\).
Exercise \(\PageIndex{6}\)
Prove theorem \(\PageIndex{3}\).
 Hint

We know that \( \displaystyle \sum_{n=0}^{\infty }\left a_nb^n \right \) converges. This should be all set for the WeierstrassM test.
To ﬁnish the story on diﬀerentiating and integrating power series, all we need to do is show that the power series, its integrated series, and its diﬀerentiated series all have the same radius of convergence. You might not realize it, but we already know that the integrated series has a radius of convergence at least as big as the radius of convergence of the original series. Speciﬁcally, suppose \(f(x) = \sum_{n=0}^{\infty }a_nx^n\) has a radius of convergence \(r\) and let \(x < r\). We know that \( \displaystyle \sum_{n=0}^{\infty }a_nx^n\)converges uniformly on an interval containing \(0\) and \(x\), and so by Corollary 8.2.2, \(\int_{t=0}^{x}f(t)dt = \sum_{n=0}^{\infty }\left (\frac{a_n}{n+1}x^{n+1} \right )\). In other words, the integrated series converges for any \(x\) with \(x < r\). This says that the radius of convergence of the integrated series must be at least \(r\).
To show that the radii of convergence are the same, all we need to show is that the radius of convergence of the diﬀerentiated series is at least as big as \(r\) as well. Indeed, since the diﬀerentiated series of the integrated series is the original, then this would say that the original series and the integrated series have the same radii of convergence. Putting the diﬀerentiated series into the role of the original series, the original series is now the integrated series and so these would have the same radii of convergence as well. With this in mind, we want to show that if \(x < r\), then \( \displaystyle \sum_{n=0}^{\infty }a_n nx^{n1}\) converges. The strategy is to mimic what we did in Theorem \(\PageIndex{1}\), where we essentially compared our series with a converging geometric series. Only this time we need to start with the diﬀerentiated geometric series.
Exercise \(\PageIndex{7}\)
Show that \( \displaystyle \sum_{n=1}^{\infty } nx^{n1}\) converges for \(x < 1\).
 Hint

We know that \( \displaystyle \sum_{k=0}^{n}x^k = \frac{x^{n+1}1}{x1}\). Diﬀerentiate both sides and take the limit as n approaches inﬁnity.
Theorem \(\PageIndex{4}\)
Suppose \( \displaystyle \sum_{n=0}^{\infty }a_nx^n\) has a radius of convergence \(r\) and let \(x < r\).Then \( \displaystyle \sum_{n=1}^{\infty }a_nnx^{n1}\) converges.
Exercise \(\PageIndex{8}\)
Prove theorem \(\PageIndex{4}\).
 Hint

Let \(b\) be a number with \(x < b < r\) and consider \(\left  a_nnx^{n1} \right  = \left  a_nb^n\cdot \frac{1}{b}\cdot n\left ( \frac{x}{b} \right )^{n1} \right \). You should be able to use the Comparison Test and Exercise \(\PageIndex{7}\).
Contributor
Eugene Boman (Pennsylvania State University) and Robert Rogers (SUNY Fredonia)