Uniform Convergence Inside the Radius

From Pointwise Convergence to Uniform Convergence

The Cauchy–Hadamard Formula and the Convergence Inside and Outside the Radius Theorem describe convergence at each individual point: a power series converges absolutely whenever the point is strictly inside its radius. But on an interval, there is a stronger question. Can one choose a single index after which the series’ partial sums approximate their limit closely at every point of that interval?

The answer is yes on every closed interval strictly inside the radius. The key is to bound all the terms at once using their values at the interval’s greatest distance from the center. The coefficient growth estimates used in the Root Criterion for a Power Series then give a geometric bound that does not depend on the point in the interval.

For the power series

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

write $S_N(x)=\sum_{n=0}^{N}c_n(x-a)^n$ for its $N$th partial sum. The radius of convergence is $R$. When $0\leq r<R$, the closed interval centered at $a$ with radius $r$ is $[a-r,a+r]$. Uniform convergence there means that a single choice of $N$ controls the approximation error at every $x$ in that interval.

Definition (Uniform Convergence of a Sequence of Functions): A sequence of functions $f_N$ on a set $E$ converges uniformly to a function $f$ on $E$ if, for every $\varepsilon>0$, there is an index $N_0$ such that $N\geq N_0$ implies $|f_N(x)-f(x)|<\varepsilon$ for every $x\in E$. The index $N_0$ may depend on $\varepsilon$, but not on $x$.

For a series, this definition is applied to its partial sums. The limit function is the pointwise sum of the series, where that sum exists. The essential distinction from pointwise convergence is the order of the choices: pointwise convergence permits the required index to depend on $x$, whereas uniform convergence requires one index to work throughout $E$.

The Uniform Convergence Theorem Inside the Radius

Fix a closed interval $[a-r,a+r]$ with $0<r<R$. Let $L=\limsup_{n\to\infty}|c_n|^{1/n}$, so the Cauchy–Hadamard Formula gives $R=1/L$, with its usual conventions when $L=0$ or $L=\infty$. The strict inequality $r<R$ implies $rL<1$. We can therefore choose a positive number $s$ such that $L<s<1/r$. This remains possible when $L=0$, since $r$ is positive and finite.

By the definition of the limsup, $|c_n|^{1/n}\leq s$ for all sufficiently large $n$. Thus $|c_n|\leq s^n$ for those indices. At every point of the interval, $|x-a|\leq r$, so each corresponding term is bounded by $(sr)^n$. Since $sr<1$, these bounds have a finite geometric sum, and their tails tend to zero independently of $x$.

Theorem (Uniform Convergence Inside the Radius): Let $\sum_{n=0}^{\infty}c_n(x-a)^n$ have radius of convergence $R$. For every finite $r$ with $0\leq r<R$, the series converges uniformly and absolutely on $[a-r,a+r]$. If $r>0$, choose $s$ with $L<s<1/r$, where $L=\limsup_{n\to\infty}|c_n|^{1/n}$, and put $q=rs$. For all sufficiently large $N$, the remainder satisfies $$ \sup_{|x-a|\leq r}\left|\sum_{n=N+1}^{\infty}c_n(x-a)^n\right| \leq \frac{q^{N+1}}{1-q}. $$

Proof. If $r=0$, the interval consists only of the center $a$. All terms with $n\geq1$ vanish there, so convergence is uniform and absolute on this one-point set. Now suppose $0<r<R$. Choose $s$ with $L<s<1/r$, and set $q=rs$, so $0<q<1$. By the definition of $L$, there is an integer $n_0$ such that $|c_n|^{1/n}\leq s$ whenever $n\geq n_0$. Equivalently, $|c_n|\leq s^n$ for those indices. Hence, for every $|x-a|\leq r$ and every $n\geq n_0$,

|c_n(x-a)^n| \leq s^n r^n =q^n.

For integers $M\geq N\geq n_0$, the finite tail is therefore bounded by

\left|\sum_{n=N+1}^{M}c_n(x-a)^n\right| \leq \sum_{n=N+1}^{M}q^n \leq \frac{q^{N+1}}{1-q}.

The final bound tends to zero as $N$ tends to infinity and does not depend on $x$. In particular, at every such $x$ the partial sums are Cauchy, so they converge in $\mathbb{R}$. Let their limit be $f(x)$. Taking $M$ to infinity in the finite-tail inequality gives

|f(x)-S_N(x)| \leq \frac{q^{N+1}}{1-q} \qquad (|x-a|\leq r,\ N\geq n_0).

Given $\varepsilon>0$, choose $N\geq n_0$ so that $q^{N+1}/(1-q)<\varepsilon$. The displayed inequality then holds with right-hand side less than $\varepsilon$ for every $x$ in the interval. This proves uniform convergence. The termwise estimate also gives

\sum_{n=n_0}^{\infty}\sup_{|x-a|\leq r}|c_n(x-a)^n| \leq \sum_{n=n_0}^{\infty}q^n <\infty.

Thus the series of absolute values is bounded by a convergent geometric series uniformly across the interval. This proves the asserted absolute convergence and the remainder bound. $\square$

When $R=\infty$, the theorem still applies on every finite interval $[a-r,a+r]$. In that case $L=0$, so for any fixed finite $r>0$, one can choose $s>0$ small enough that $rs<1$. If $R=0$, there is no positive $r<R$; the theorem only applies at the center, where the power-series convergence at $x=a$ is already known.

Worked Examples: Using a Uniform Tail Bound

Worked Example: A Polynomial Factor on a Smaller Interval

Consider

\sum_{n=1}^{\infty}\frac{n^2}{6^n}(x+1)^n.

The coefficient is $c_n=n^2/6^n$. Since $(n^2)^{1/n}\to1$, its coefficient-root limsup is $L=1/6$, and the radius is $R=6$. Take the interval $[-3,1]$, which is centered at $a=-1$ with $r=2$. Choose $s=1/5$. Then $L=1/6<1/5<1/2=1/r$, and $q=rs=2/5$. For all sufficiently large $n$, $|c_n|^{1/n}\leq1/5$. Consequently, on the whole interval,

\left|\frac{n^2}{6^n}(x+1)^n\right| \leq \left(\frac25\right)^n.

If $N$ is large enough for this estimate to apply to every $n\geq N+1$, the remainder after $N$ terms obeys

\sup_{-3\leq x\leq1}|S(x)-S_N(x)| \leq \sum_{n=N+1}^{\infty}\left(\frac25\right)^n =\frac{(2/5)^{N+1}}{1-2/5} =\frac{5}{3}\left(\frac25\right)^{N+1}.

The right-hand side tends to zero. The same $N$ therefore approximates the sum to any prescribed accuracy at every point of $[-3,1]$.

Worked Example: An Infinite Radius and Any Fixed Bounded Interval

Consider the series centered at $a=2$,

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

Its coefficients satisfy $c_n=1/n!$ and, for $n\geq0$, $c_{n+1}/c_n=1/(n+1)$. The Ratio Criterion for the Radius of a Power Series gives $R=\infty$. Take any finite $r>0$. For a direct geometric bound, choose an integer $n_0$ with $n_0+1\geq 2r$. If $n\geq n_0$, then

\frac{r^{n+1}/(n+1)!}{r^n/n!} =\frac{r}{n+1} \leq\frac12.

Thus the successive upper bounds $r^n/n!$ decrease at least geometrically after $n_0$. For every $|x-2|\leq r$, the absolute value of the $n$th term is at most $r^n/n!$. For $N\geq n_0$, it follows that

\sup_{|x-2|\leq r}\left|\sum_{n=N+1}^{\infty}\frac{(x-2)^n}{n!}\right| \leq \sum_{n=N+1}^{\infty}\frac{r^n}{n!} \leq \frac{r^{N+1}}{(N+1)!}\sum_{j=0}^{\infty}\left(\frac12\right)^j =\frac{2r^{N+1}}{(N+1)!}.

For the second inequality, each successive term after $r^{N+1}/(N+1)!$ is at most half its predecessor, by the ratio estimate. The final bound tends to zero, so the series converges uniformly on every fixed finite interval around $2$. The interval may be chosen as large as desired, but it must remain bounded for this argument.

Worked Example: A Geometric Power Series on a Smaller Interval

For $\sum_{n=0}^{\infty}x^n$, the radius is $1$. On $[-1/2,1/2]$, the absolute value of the $n$th term is at most $(1/2)^n$. The finite geometric identity gives, for $x\neq1$,

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

The multiplication verifies the identity. The sum is $1/(1-x)$ for $|x|<1$, and its remainder on $[-1/2,1/2]$ is

\left|\frac{1}{1-x}-\sum_{n=0}^{N}x^n\right| =\left|\frac{x^{N+1}}{1-x}\right| \leq \frac{(1/2)^{N+1}}{1-1/2} =2^{-N}.

Here $|1-x|\geq1-|x|\geq1/2$, which justifies the inequality. Since $2^{-N}\to0$, the convergence is uniform on this interval. This example gives the exact remainder as well as a uniform bound.

Why the Whole Open Interval Is Different

Uniform convergence on every smaller closed interval does not imply uniform convergence on the entire open interval $(a-R,a+R)$. In a smaller interval, the distance from the center is bounded by some $r<R$, leaving a fixed gap between $r$ and the radius. That gap allows the geometric factor $q=rs$ to be strictly less than one. As the interval expands toward the boundary, this uniform margin may disappear.

For the geometric series $\sum_{n=0}^{\infty}x^n$, the pointwise sum on $(-1,1)$ is $1/(1-x)$. Its remainder after the $N$th partial sum is $x^{N+1}/(1-x)$. For any fixed $N$, this remainder is unbounded as $x$ approaches $1$ from below. Indeed, for $0<x<1$, the numerator tends to $1$ while the denominator tends to $0$. Therefore no single $N$ makes the remainder small throughout $(-1,1)$: the convergence is not uniform there.

A related caution is that a geometric estimate valid for a fixed smaller interval should not be used with its radius replaced by the full radius. The proof requires $q<1$. At the boundary, the corresponding factor can equal one, and the geometric tail estimate no longer tends to zero. Endpoint convergence must be analyzed separately, as in the Endpoint Classification of the Interval of Convergence.

Choose a closed interval strictly inside the radius.
Write it as $[a-r,a+r]$ with $r<R$.

Turn coefficient growth into a term bound.
Choose $s$ above the coefficient-root limsup but with $rs<1$. Eventually, $|c_n|\leq s^n$.

Bound the whole tail geometrically.
Since $|x-a|\leq r$, each sufficiently late term is at most $(rs)^n$, uniformly in $x$.

Use the bound to choose one index.
The geometric tail tends to zero, so a single partial sum approximates the series throughout the closed interval.

Takeaway: A power series converges uniformly and absolutely on every closed interval whose endpoints lie strictly inside its radius. Its tail has a geometric bound independent of the point in that interval. This local uniform conclusion does not, by itself, give uniform convergence on the entire open interval of convergence.

Check Your Understanding

Use the uniform convergence theorem and its estimates to answer the following questions.

What is the difference between pointwise and uniform convergence of the partial sums on a set?
Why can one choose $s$ with $L<s<1/r$ when $0<r<R$?
If $r=2$ and a suitable coefficient bound gives $s=1/4$, what is the geometric factor $q$, and does it give a convergent tail bound?
Why does the uniform convergence theorem still apply to every finite interval when a power series has infinite radius?
For $\sum_{n=0}^{\infty}x^n$, why does the remainder formula rule out uniform convergence on $(-1,1)$?

Uniform Convergence Inside the Radius

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