Power Series as Function Sequences

A Power Series Is a Sequence of Functions

A power series can be studied not only as a sum at each point, but also as a sequence of functions. This viewpoint makes the distinction between pointwise and uniform convergence especially clear. At each stage, a partial sum is a function; convergence of the series means that this sequence of functions approaches its sum in the appropriate sense.

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

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

Its \(N\)th partial sum and its \(n\)th term function are

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

For each fixed \(N\), \(S_N\) is a polynomial, hence a function of \(x\). The series converges at a point \(x\) precisely when the numerical sequence \(S_N(x)\) converges. It converges uniformly on a set \(E\) when the sequence of functions \(S_N\) converges uniformly there: one index must make the error small for every \(x\in E\).

Definition (Partial-Sum Functions): For a power series \(\sum_{n=0}^{\infty}c_n(x-a)^n\), the partial-sum functions are \(S_N(x)=\sum_{n=0}^{N}c_n(x-a)^n\). The series has a sum function \(S\) on \(E\) if \(S_N(x)\to S(x)\) for every \(x\in E\). The convergence is uniform on \(E\) if, for every \(\varepsilon>0\), there is an \(N_0\) such that \(N\geq N_0\) implies \(|S_N(x)-S(x)|<\varepsilon\) for every \(x\in E\).

The individual term functions \(u_n\) are not the partial sums. They record the change from one partial sum to the next:

S_n(x)-S_{n-1}(x)=u_n(x) \qquad(n\geq1).

This identity is useful, but it does not make convergence of the term functions equivalent to convergence of the partial sums. A sequence of terms can tend uniformly to zero while the sums still fail to converge even pointwise. Conversely, uniform convergence of the partial sums does force the terms to tend uniformly to zero, as we will prove below.

Uniform Convergence and the Uniform Cauchy Criterion

The Uniform Convergence Inside the Radius Theorem gives an important example of uniform convergence of partial-sum functions: on every closed interval strictly inside the radius, those functions converge uniformly. A general way to test uniform convergence is to ask whether the partial sums become uniformly close to one another. This is the function-sequence analogue of the Cauchy Criterion for Series.

Theorem (Uniform Cauchy Criterion for Partial Sums): Let \(E\) be a set, and let \(S_N\) be the partial-sum functions of a series on \(E\). The sequence \((S_N)\) converges uniformly on \(E\) if and only if, for every \(\varepsilon>0\), there is an index \(N_0\) such that, for all \(m,n\geq N_0\) and all \(x\in E\), \(|S_m(x)-S_n(x)|<\varepsilon\).

Proof. Suppose first that \(S_N\) converges uniformly to \(S\) on \(E\). Given \(\varepsilon>0\), uniform convergence gives an index \(N_0\) such that

|S_k(x)-S(x)|<\frac{\varepsilon}{2} \qquad(k\geq N_0,\ x\in E).

For \(m,n\geq N_0\) and \(x\in E\), the triangle inequality gives

|S_m(x)-S_n(x)| \leq |S_m(x)-S(x)|+|S_n(x)-S(x)| <\varepsilon.

Thus the partial sums are uniformly Cauchy.

Conversely, suppose that the stated uniform Cauchy condition holds. Fix \(x\in E\). For every \(\varepsilon>0\), the condition implies that \(S_N(x)\) is a Cauchy sequence of real numbers. By the Cauchy Criterion for Sequences in \(\mathbb{R}\), it converges. Define \(S(x)=\lim_{N\to\infty}S_N(x)\). This defines a function \(S\) at every point of \(E\).

Now fix \(\varepsilon>0\), and apply the uniform Cauchy condition with \(\varepsilon/2\). There is an \(N_0\) such that

|S_m(x)-S_n(x)|<\frac{\varepsilon}{2} \qquad(m,n\geq N_0,\ x\in E).

Fix \(n\geq N_0\) and \(x\in E\), and let \(m\) tend to infinity. Since \(S_m(x)\to S(x)\), it follows that \(|S(x)-S_n(x)|\leq\varepsilon/2<\varepsilon\). This bound holds for every \(x\in E\), so \(S_N\) converges uniformly to \(S\). \(\square\)

For partial sums, the difference between two stages is a finite block of terms. If \(m>n\), then

S_m(x)-S_n(x)=\sum_{k=n+1}^{m}c_k(x-a)^k.

Therefore, the uniform Cauchy criterion asks whether every sufficiently late finite block has small absolute value at every point of the set. It does not require an explicit formula for the infinite sum. This is particularly useful when the sum function is unknown or difficult to calculate.

Uniform Convergence Forces the Terms to Vanish Uniformly

For numerical series, convergence requires the terms to tend to zero. For a series of functions, uniform convergence yields a stronger version of that necessary condition: the term functions must tend to zero uniformly. This condition can rule out uniform convergence, although satisfying it is not enough to prove uniform convergence.

Theorem (Uniform Convergence Forces Uniformly Vanishing Terms): Suppose the partial-sum functions \(S_N\) of a series converge uniformly to \(S\) on \(E\). If \(u_n=S_n-S_{n-1}\) for \(n\geq1\), then \(u_n\to0\) uniformly on \(E\).

Proof. Let \(\varepsilon>0\). Uniform convergence gives an index \(N_0\) such that, for every \(k\geq N_0\) and every \(x\in E\),

|S_k(x)-S(x)|<\frac{\varepsilon}{2}.

For \(n\geq N_0+1\), both \(n\) and \(n-1\) are at least \(N_0\). Using \(u_n(x)=S_n(x)-S_{n-1}(x)\), we obtain, for every \(x\in E\),

|u_n(x)| \leq |S_n(x)-S(x)|+|S_{n-1}(x)-S(x)| <\varepsilon.

The same index \(N_0+1\) works for every \(x\in E\), which is precisely uniform convergence of \(u_n\) to zero. \(\square\)

The converse is false: uniformly small terms need not have partial sums that converge, even pointwise. The next example demonstrates this failure at an endpoint.

Worked Examples: Reading the Function Sequence

Worked Example: Partial Sums on a Closed Interval

Consider the power series centered at \(a=1\),

\sum_{n=0}^{\infty}\left(\frac{x-1}{4}\right)^n.

Its term functions are \(u_n(x)=((x-1)/4)^n\), and its \(N\)th partial-sum function is \(S_N(x)=\sum_{n=0}^{N}((x-1)/4)^n\). On \(E=[0,2]\), one has \(|x-1|\leq1\), so \(|(x-1)/4|\leq1/4\). The finite geometric identity gives, with \(t=(x-1)/4\),

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

The second identity verifies the formula, and \(1-t\neq0\) throughout \(E\) because \(|t|\leq1/4\). The limit function is \(S(x)=1/(1-t)\), since \(t^{N+1}\to0\) at each point. Moreover, for every \(x\in E\),

|S(x)-S_N(x)| =\left|\frac{t^{N+1}}{1-t}\right| \leq\frac{(1/4)^{N+1}}{1-1/4} =\frac{4}{3}\left(\frac14\right)^{N+1}.

Here \(|1-t|\geq1-|t|\geq3/4\). The final bound is independent of \(x\) and tends to zero. Thus the partial-sum functions converge uniformly on \([0,2]\). This example shows how a pointwise geometric identity can provide a uniform error estimate for a whole set.

Worked Example: Uniformly Vanishing Terms but Divergent Partial Sums

Consider

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

on \(E=[0,1]\). The term function is \(u_n(x)=x^n/n\). Since \(0\leq x\leq1\),

0\leq |u_n(x)|=\frac{x^n}{n}\leq\frac{1}{n}.

The bound \(1/n\) tends to zero independently of \(x\), so the term functions converge uniformly to zero on \(E\). But at \(x=1\), the partial sums are

S_N(1)=\sum_{n=1}^{N}\frac{1}{n}.

These are the partial sums of the \(p\)-series with \(p=1\), which diverges by the p-Series Convergence Criterion. Therefore the power series does not even converge pointwise on all of \(E\), and in particular it cannot converge uniformly there. Uniformly vanishing terms are necessary for uniform convergence, but they are not sufficient.

Worked Example: Applying the Uniform Cauchy Criterion

Consider the series

\sum_{n=1}^{\infty}\frac{(x-3)^n}{n^2 5^n}

on \(E=[2,4]\). For \(x\in E\), \(|x-3|\leq1\). If \(m>n\geq N\geq1\), then

|S_m(x)-S_n(x)| \leq\sum_{k=n+1}^{m}\frac{|x-3|^k}{k^2 5^k} \leq\sum_{k=n+1}^{m}\left(\frac15\right)^k \leq\frac{(1/5)^{n+1}}{1-1/5}.

The last expression is at most \((1/5)^{N+1}/(1-1/5)\), which tends to zero as \(N\) tends to infinity and does not depend on \(x\), \(m\), or \(n\). Given \(\varepsilon>0\), choose \(N\) large enough that this bound is less than \(\varepsilon\). Then all partial sums with indices at least \(N\) are within \(\varepsilon\) of one another at every point of \(E\). The Uniform Cauchy Criterion for Partial Sums proves that the series converges uniformly on \([2,4]\), without first requiring a formula for its sum.

What the Function-Sequence Viewpoint Clarifies

A power series supplies two related sequences: its term functions \(u_n\) and its partial-sum functions \(S_N\). The terms describe individual increments, while the partial sums describe the approximations whose convergence defines the sum function. Confusing these sequences can lead to an invalid conclusion: proving \(u_n\to0\) uniformly does not prove that \(S_N\) converges uniformly.

The uniform Cauchy criterion provides a test directly in terms of finite blocks of terms. It is often more practical than estimating the difference between partial sums and a sum function that has not yet been identified. The Uniform Convergence Inside the Radius Theorem is one powerful way to meet this criterion on a closed interval strictly inside the radius. The estimates there control the tails independently of the point.

This viewpoint also prepares for studying properties of the sum function. The polynomials \(S_N\) are simple approximating functions, and uniform convergence controls how closely they approximate their limit across an entire set. The next step is to investigate what such control implies for continuity.

Takeaway: A power series is a sequence of partial-sum functions, and convergence of the series means convergence of that sequence. Uniform convergence is equivalent to a uniform Cauchy condition on finite blocks of terms; it also forces the individual term functions to tend uniformly to zero, though that necessary condition alone does not ensure convergence.

Check Your Understanding

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

For a power series, what is the difference between a term function and a partial-sum function?
State the uniform Cauchy condition for the partial-sum functions \(S_N\).
Why does uniform convergence of \(S_N\) imply that \(u_n=S_n-S_{n-1}\) tends uniformly to zero?
On \([0,1]\), why do the term functions \(x^n/n\) tend uniformly to zero, and why does their series nevertheless fail to converge at every point of that set?
In the third worked example, which bound makes the Cauchy estimate independent of \(x\), \(m\), and \(n\)?