Pointwise Convergence of Function Series

Convergence One Input at a Time

The previous tutorial treated uniform convergence of a function series by requiring its tails to be small across the entire domain at once. Pointwise convergence asks for less: fix an input $x$, and ask whether the resulting numerical series converges. The index after which the tails are small may depend on the chosen input.

Let $E$ be a nonempty set, and let $f_n:E\to\mathbb{R}$. The $N$th partial sum is the function $S_N(x)=\sum_{n=1}^{N}f_n(x)$. At a fixed $x$, the values $f_1(x),f_2(x),\ldots$ form an ordinary numerical series. Pointwise convergence of the function series means that this numerical series converges for every $x\in E$.

Definition: The series $\sum_{n=1}^{\infty}f_n(x)$ converges pointwise on $E$ if, for every $x\in E$, the sequence of partial sums $S_N(x)=\sum_{n=1}^{N}f_n(x)$ converges in $\mathbb{R}$. Its pointwise sum is the function $S:E\to\mathbb{R}$ given by $S(x)=\lim_{N\to\infty}S_N(x)$.

The quantifiers are important. Pointwise convergence says that for each $x$ and each $\varepsilon>0$, there is an index $N$ that works at that $x$. It does not require the same $N$ to work for every input. In symbols, the index can depend on both $x$ and $\varepsilon$.

\text{For every }x\in E\text{ and }\varepsilon>0,\text{ there is }N=N(x,\varepsilon)\text{ such that }|S_n(x)-S(x)|<\varepsilon\text{ for all }n\geq N.

For a uniformly convergent series, by contrast, the index can depend on $\varepsilon$ but not on $x$. Pointwise convergence is therefore a separate, input-by-input question about ordinary series, not a uniform estimate on the domain.

A Cauchy Criterion for Pointwise Convergence

The pointwise criterion for a series applies the numerical Cauchy criterion separately at each input. It is useful because it tests tails without requiring us to know the sum in advance. The index in the criterion may vary with the input.

Theorem (Pointwise Cauchy Criterion for a Series): The series $\sum_{n=1}^{\infty}f_n(x)$ converges pointwise on $E$ if and only if, for every $x\in E$ and every $\varepsilon>0$, there is an integer $N=N(x,\varepsilon)\geq0$ such that for all integers $q\geq p\geq N$, $$ \left|\sum_{n=p+1}^{q}f_n(x)\right|<\varepsilon. $$

Proof. Fix $x\in E$. For the partial sums at this input, $$ S_q(x)-S_p(x)=\sum_{n=p+1}^{q}f_n(x). $$ If the series converges at $x$, then $S_N(x)$ is a convergent, hence Cauchy, sequence of real numbers. Given $\varepsilon>0$, there is an $N$ such that $|S_q(x)-S_p(x)|<\varepsilon$ whenever $q\geq p\geq N$. The displayed identity gives the required tail bound.

Conversely, suppose the tail condition holds at this fixed $x$. The same identity shows that $(S_N(x))$ is a Cauchy sequence of real numbers. Completeness of $\mathbb{R}$ implies that it converges. Since the condition holds for every $x\in E$, the series converges pointwise on $E$. The argument is applied separately at each input, so it does not produce a single index that works on all of $E$. $\square$

Worked Example: A Telescoping Series with a Pointwise Sum

For $x\in[0,\infty)$, define $$ f_n(x)=\frac{1}{n+x}-\frac{1}{n+1+x},\qquad n\geq1. $$ The partial sum through $N$ telescopes: $$ S_N(x)=\sum_{n=1}^{N}\left(\frac{1}{n+x}-\frac{1}{n+1+x}\right) =\frac{1}{1+x}-\frac{1}{N+1+x}. $$ For each fixed $x\geq0$, the second term tends to zero as $N\to\infty$. Thus the series converges pointwise to $S(x)=1/(1+x)$.

This calculation also gives the exact remainder: $$ \left|S(x)-S_N(x)\right|=\frac{1}{N+1+x}. $$ At a fixed $x$, this tends to zero. In fact, here the bound is at most $1/(N+1)$ for every $x\geq0$, so the convergence is also uniform. The pointwise calculation itself, however, only needed $x$ to be fixed.

A Necessary Condition on the Terms

If a series converges at an input, its terms must tend to zero at that input. This necessary condition follows by subtracting consecutive partial sums. As with the definition of pointwise convergence, the resulting limit is pointwise; the terms need not tend uniformly to zero.

Proposition (Terms of a Pointwise Convergent Series Vanish Pointwise): If $\sum_{n=1}^{\infty}f_n(x)$ converges pointwise on $E$, then $f_n(x)\to0$ for every $x\in E$.

Proof. Fix $x\in E$, and let $S(x)$ be the sum. For $n\geq2$, the definition of partial sums gives $f_n(x)=S_n(x)-S_{n-1}(x)$. Since both $S_n(x)$ and $S_{n-1}(x)$ tend to $S(x)$, the triangle inequality yields $$ |f_n(x)|\leq |S_n(x)-S(x)|+|S_{n-1}(x)-S(x)|\longrightarrow0. $$ This holds for each fixed $x$, proving the claim. $\square$

The converse is false: terms tending to zero do not guarantee that the series converges. For example, the terms of the harmonic series tend to zero, but its partial sums are unbounded. Thus the pointwise Cauchy criterion checks the entire tail, while the condition on individual terms checks only one summand at a time.

Worked Example: A Pointwise Limit with No Uniform Tail Control

On $E=(0,1]$, define $$ f_n(x)=x(1-x)^{n-1},\qquad n\geq1. $$ For each fixed $x\in(0,1]$, the finite geometric-sum identity gives $$ S_N(x)=\sum_{n=1}^{N}x(1-x)^{n-1}=1-(1-x)^N. $$ Because $0\leq1-x<1$, the term $(1-x)^N$ tends to zero. Therefore the series converges pointwise to the constant function $S(x)=1$.

But the error after $N$ terms is $$ |S(x)-S_N(x)|=(1-x)^N. $$ For every fixed $N$, this error approaches $1$ as $x$ approaches $0$ through positive values. Consequently, $$ \sup_{x\in(0,1]}|S(x)-S_N(x)|=1 $$ for every $N$: the values are less than $1$, but can be made arbitrarily close to it. The partial sums do not converge uniformly, even though they converge at every point of the domain. The small errors at each fixed input do not provide one index that works across the domain.

Convergence Can Depend on the Input

A function series need not converge at every point of a proposed domain. Its convergence set is the collection of inputs at which the associated numerical series converges. The next example shows why a boundary point must be checked separately from inputs where a geometric comparison applies.

Worked Example: A Series That Fails at One Endpoint

Consider $$ \sum_{n=1}^{\infty}\frac{x^n}{n} $$ on $[-1,1]$. If $|x|<1$, then $$ \left|\frac{x^n}{n}\right|\leq |x|^n, $$ and the geometric series $\sum_{n=1}^{\infty}|x|^n$ converges. The comparison test therefore gives absolute convergence for every $-1<x<1$.

At $x=-1$, the series becomes $\sum_{n=1}^{\infty}(-1)^n/n$, which converges by the alternating-series test: its positive magnitudes $1/n$ decrease to zero. At $x=1$, it becomes the harmonic series $\sum_{n=1}^{\infty}1/n$, which diverges. Hence the function series converges at precisely the points in $[-1,1)$. A pointwise sum is defined on that convergence set, but not on all of $[-1,1]$.

For $-1<x<1$, the logarithm-series formula gives the sum $-\ln(1-x)$. At $x=-1$, the sum is $-\ln 2$, also given by the endpoint value of the alternating logarithm series. No finite sum exists at $x=1$. This illustrates why the pointwise definition asks for convergence at each input in the stated domain rather than assuming that convergence at nearby inputs settles the boundary case.

Pointwise and Uniform Statements

The difference between pointwise and uniform convergence is a difference in quantifier order. In pointwise convergence, one fixes $x$ before choosing the index. In uniform convergence, the index must be chosen before considering which $x$ will be tested. The series in the previous example shows that a domain may include a point at which convergence fails; the preceding geometric example shows that convergence at every point still need not be uniform.

Fix an input.
For pointwise convergence, choose $x\in E$ and consider the numerical partial sums $S_N(x)$.

Check the series at that input.
Use ordinary numerical-series tools, or verify the pointwise Cauchy criterion for its finite tails.

Repeat across the domain.
The series converges pointwise on $E$ only if the check succeeds for every $x\in E$; the index may differ from point to point.

A common pitfall is to establish that each numerical series converges and then infer uniform convergence. That conclusion needs additional control, such as the uniform Cauchy criterion for a series or a suitable uniform convergence test. Another pitfall is to test only whether $f_n(x)\to0$: this is necessary, but the harmonic-series example shows why it does not control sums of many small terms.

Takeaway: A function series converges pointwise when its numerical series converges at every input. Its finite tails must become small at each fixed input, with an index that may depend on that input. The terms must tend to zero pointwise, but only control of the full tails decides whether the series converges.

Check Your Understanding

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

In the definition of pointwise convergence, which quantities may the index depend on?
How does the Pointwise Cauchy Criterion for a Series express convergence without referring to the sum?
Why does convergence of a series at each input force its terms to tend to zero there?
Why does the series $\sum_{n=1}^{\infty}x(1-x)^{n-1}$ converge pointwise on $(0,1]$ but not uniformly?
At which point of $[-1,1]$ does $\sum_{n=1}^{\infty}x^n/n$ diverge, and what numerical series does it become there?