Continuity of Power Series

From Uniform Convergence to Continuity

In the previous tutorial, a power series was viewed as a sequence of partial-sum functions. Each partial sum is a polynomial, so it is continuous. The key question is whether continuity passes from these approximating polynomials to their limit. It does when the convergence is uniform: on a set where the approximation error is uniformly small, continuity of one suitable partial sum controls the limit nearby.

We first state continuity for a function defined on an arbitrary set \(E\subseteq\mathbb{R}\). This includes intervals and closed intervals, which will be used when we apply the result to power series.

Definition (Continuity on a Set): Let \(f:E\to\mathbb{R}\), and let \(x_0\in E\). The function \(f\) is continuous at \(x_0\) relative to \(E\) if, for every \(\varepsilon>0\), there is a \(\delta>0\) such that \(x\in E\) and \(|x-x_0|<\delta\) imply \(|f(x)-f(x_0)|<\varepsilon\). The function is continuous on \(E\) if it is continuous at every point of \(E\).

The phrase “relative to \(E\)” matters when \(x_0\) is an endpoint of an interval: only points belonging to \(E\) are tested. In the power series application, we will choose \(x_0\) strictly inside an interval, so nearby points on both sides are available.

Theorem (Uniform Limits of Continuous Functions Are Continuous): Let \(E\subseteq\mathbb{R}\), and suppose each function \(f_n:E\to\mathbb{R}\) is continuous on \(E\). If \(f_n\) converges uniformly on \(E\) to \(f\), then \(f\) is continuous on \(E\).

Proof. Fix \(x_0\in E\), and let \(\varepsilon>0\). Uniform convergence gives an index \(N\) such that

|f_N(x)-f(x)|<\frac{\varepsilon}{3} \qquad\text{for every }x\in E.

The function \(f_N\) is continuous at \(x_0\). Thus there is a \(\delta>0\) such that, whenever \(x\in E\) and \(|x-x_0|<\delta\),

|f_N(x)-f_N(x_0)|<\frac{\varepsilon}{3}.

For such an \(x\), insert the values of \(f_N\) between \(f(x)\) and \(f(x_0)\). The triangle inequality gives

\begin{aligned} |f(x)-f(x_0)| &\leq |f(x)-f_N(x)| +|f_N(x)-f_N(x_0)| +|f_N(x_0)-f(x_0)|\\ &<\frac{\varepsilon}{3} +\frac{\varepsilon}{3} +\frac{\varepsilon}{3} =\varepsilon. \end{aligned}

Therefore \(f\) is continuous at \(x_0\). Since \(x_0\) was arbitrary, \(f\) is continuous on \(E\). \(\square\)

The proof uses one index \(N\) to control the approximation error at every point of \(E\), and then uses continuity of that single function \(f_N\) near \(x_0\). If convergence were only pointwise, the approximation index could depend on the point, and this argument would not provide a uniform error bound nearby.

Continuity of a Power Series Inside Its Radius

Consider a power series centered at \(a\),

\sum_{n=0}^{\infty}c_n(x-a)^n, \qquad S_N(x)=\sum_{n=0}^{N}c_n(x-a)^n.

Each \(S_N\) is a polynomial and hence continuous. The Uniform Convergence Inside the Radius Theorem says that on every closed interval \([a-r,a+r]\) with \(0\leq r<R\), where \(R\) is the radius of convergence, these partial sums converge uniformly. Applying the uniform limit theorem on such an interval proves continuity there. To handle a particular point, choose an interval of this kind that contains it in its interior.

Theorem (Continuity of a Power Series Inside Its Radius): Let \(R\) be the radius of convergence of \(\sum_{n=0}^{\infty}c_n(x-a)^n\). Its sum function is continuous at every \(x_0\) satisfying \(|x_0-a|<R\). In particular, if \(R>0\), the sum function is continuous on the open interval \((a-R,a+R)\).

Proof. Fix \(x_0\) with \(|x_0-a|<R\). Choose a finite number \(r\) such that

$$ |x_0-a|<r<R. $$

If \(R=\infty\), choose any finite \(r>|x_0-a|\). The point \(x_0\) lies in the interior of \(I=[a-r,a+r]\). By the Uniform Convergence Inside the Radius Theorem, the partial-sum functions \(S_N\) converge uniformly on \(I\). Each \(S_N\) is continuous on \(I\), so the Uniform Limits of Continuous Functions Are Continuous Theorem shows that their limit is continuous on \(I\). In particular, the power series sum is continuous at \(x_0\). Since \(x_0\) was any point strictly inside the radius, the result follows. \(\square\)

This is a local argument. It does not require the partial sums to converge uniformly on the entire open interval \((a-R,a+R)\). For each point, it is enough to find one closed interval around that point whose radius from the center \(a\) is still strictly less than \(R\).

Worked Examples: Applying the Continuity Theorem

Worked Example: A Geometric Power Series

Consider the power series centered at \(2\),

\sum_{n=0}^{\infty}\left(\frac{x-2}{3}\right)^n.

Its radius of convergence is \(3\), so the Continuity of a Power Series Inside Its Radius Theorem guarantees that its sum is continuous for \(|x-2|<3\), or \(-1<x<5\). We can also identify the sum. Put \(t=(x-2)/3\). The finite geometric-sum identity is

\sum_{n=0}^{N}t^n=\frac{1-t^{N+1}}{1-t}, \qquad (1-t)\sum_{n=0}^{N}t^n=1-t^{N+1}.

The second equation verifies the identity. When \(|t|<1\), \(t^{N+1}\to0\), so the series sums to \(1/(1-t)\). Substituting \(t=(x-2)/3\) gives

\frac{1}{1-(x-2)/3}=\frac{3}{5-x}.

For \(-1<x<5\), the denominator \(5-x\) is nonzero. Thus the explicit sum is continuous throughout the open interval, in agreement with the theorem.

Worked Example: A Power Series with Infinite Radius

Consider

\sum_{n=0}^{\infty}\frac{(x+1)^n}{n!}.

For any fixed \(x\), the ratio of the absolute values of consecutive terms, once the terms are nonzero, is

\frac{|x+1|^{n+1}/(n+1)!}{|x+1|^n/n!} =\frac{|x+1|}{n+1}\longrightarrow0.

If \(x=-1\), all terms after the constant term are zero; otherwise the Ratio Test applies directly. In either case, the series converges for every real \(x\), so its radius of convergence is infinite. The continuity theorem therefore shows that its sum function is continuous on all of \(\mathbb{R}\). More locally, given any point \(x_0\), choose a finite \(r>|x_0+1|\). The partial sums converge uniformly on \([-1-r,-1+r]\), and their limit is continuous there. No explicit formula for the sum is needed to reach this conclusion.

Worked Example: A Finite Radius and Its Endpoints

Consider

\sum_{n=1}^{\infty}\frac{x^n}{n}.

The coefficient root limit is \(\lim_{n\to\infty}(1/n)^{1/n}=1\), so the radius of convergence is \(1\). Its sum function is therefore continuous on \((-1,1)\). The theorem does not, by itself, determine what happens at \(x=-1\) or \(x=1\), because neither endpoint lies strictly inside the radius.

The endpoint series can be checked separately. At \(x=1\), it becomes \(\sum_{n=1}^{\infty}1/n\), which diverges by the \(p\)-Series Convergence Criterion with \(p=1\). At \(x=-1\), it becomes \(\sum_{n=1}^{\infty}(-1)^n/n\), which converges by the Alternating Series Test. Thus this series has a value at one endpoint but not the other. Endpoint convergence is a separate question from continuity inside the radius.

Why Uniformity Matters—and What the Theorem Does Not Say

The uniformity in the continuity theorem cannot be omitted. For example, on \([0,1]\), the continuous functions \(f_n(x)=x^n\) converge pointwise to the function \(f\) given by \(f(x)=0\) for \(0\leq x<1\) and \(f(1)=1\). This limit is not continuous at \(1\): values \(x<1\) can be chosen arbitrarily close to \(1\), yet \(f(x)=0\) while \(f(1)=1\). The convergence is not uniform. Indeed, for every \(n\), taking \(x\) sufficiently close to \(1\) makes \(x^n\) arbitrarily close to \(1\), so the difference from \(f(x)=0\) on points below \(1\) cannot be bounded by one small error for all \(x\).

For power series, the available uniform convergence is local to intervals strictly inside the radius. That is exactly enough to prove continuity at every interior point: each such point has room for a smaller closed interval around it. At a boundary point, no such interval can be chosen while keeping its radius strictly less than \(R\). The endpoint examples above illustrate why convergence there must be examined separately.

A further distinction is that continuity on an open interval does not assert uniform continuity on that whole interval. The theorem proves continuity at each point. Its method gives a uniform approximation on a suitable smaller closed interval around that point, not necessarily a single approximation estimate or continuity bound over the entire interval of convergence.

Takeaway: The partial sums of a power series are continuous polynomials. Their uniform convergence on every closed interval strictly inside the radius makes the sum continuous at every point strictly inside that radius. Boundary behavior requires separate analysis.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

In the proof that a uniform limit of continuous functions is continuous, where is uniform convergence used?
Why can a point \(x_0\) strictly inside a power series’ radius be placed in a closed interval on which the partial sums converge uniformly?
What does the Continuity of a Power Series Inside Its Radius Theorem conclude when the radius is infinite?
For the series \(\sum_{n=1}^{\infty}x^n/n\), what happens at each endpoint of its interval of convergence?
Why does pointwise convergence of continuous functions alone not guarantee a continuous limit?

What You'll Learn