In short

A sequence of functions f_1, f_2, f_3, \ldots is a list where every entry is itself a function. As the index grows, the functions can approach a limit function f — but in two distinct ways. Pointwise convergence means that at each fixed point x, the numbers f_n(x) approach f(x). Uniform convergence is stronger: the whole function f_n gets close to f everywhere at once, at a rate independent of x. The distinction matters because uniform convergence preserves continuity, integrability, and differentiability, while pointwise convergence does not. A power series \sum a_n x^n is an important example — an infinite sum of functions that converges inside its radius of convergence.

You have worked with sequences of numbers: 1, 1/2, 1/4, 1/8, \ldots approaching 0. You have worked with infinite series of numbers: 1 + 1/2 + 1/4 + \cdots summing to 2. Now shift one level up. What if every entry in the sequence is not a number but a function?

Here is an example. Define f_n(x) = x^n for each positive integer n. This gives you an infinite list of functions:

f_1(x) = x, \quad f_2(x) = x^2, \quad f_3(x) = x^3, \quad \ldots

At every fixed value of x, this list produces a sequence of numbers. For x = 1/2, you get 1/2, 1/4, 1/8, \ldots — approaching 0. For x = 1, you get 1, 1, 1, \ldots — constant at 1. For x = 1.1, you get 1.1, 1.21, 1.331, \ldots — growing without bound.

So for any single x, you know what happens to f_n(x) as n grows. But what about the functions themselves, viewed as whole objects? Does the sequence of functions approach some limiting function? In what sense?

The answer depends sharply on what "approach" means, and getting the definition right turns out to matter enormously. There are two main notions of convergence for sequences of functions, and the gap between them is the subject of the first half of this discussion. The second half takes the weaker structure and applies it to power series — the infinite-polynomial machinery that expresses sine, cosine, exponential, and logarithm in terms of their derivatives.

Pointwise convergence: the first definition

The easiest thing you can mean by "the functions f_n approach the function f" is this: at every single x, the numerical sequence f_n(x) approaches f(x). That is pointwise convergence.

Pointwise convergence

A sequence of functions f_n, each defined on a set D, converges pointwise to a function f on D if, for every x \in D,

\lim_{n \to \infty} f_n(x) = f(x).

Expanded via epsilon: for each x \in D and each \varepsilon > 0, there exists a natural number N (depending on both x and \varepsilon) such that n \geq N implies |f_n(x) - f(x)| < \varepsilon.

The crucial phrase is "depending on both x and \varepsilon." At one point of the domain, a small N might be enough; at another point, you might need N to be a million. Pointwise convergence allows this — it only demands that some N exists at each point separately.

Apply this to f_n(x) = x^n on the interval [0, 1].

So the pointwise limit is

f(x) = \begin{cases} 0 & \text{if } 0 \leq x < 1 \\ 1 & \text{if } x = 1. \end{cases}

Something strange has happened. Every single function f_n(x) = x^n is continuous on [0, 1]. But the limit function f has a jump at x = 1 — it goes from 0 (just below 1) to 1 (at 1). A sequence of continuous functions has converged pointwise to a discontinuous function. Continuity was not preserved by the limit.

x to the n for several values of n on the unit intervalFour curves of y equals x to the n for n equals 1, 2, 5, and 20 drawn on the interval from 0 to 1. As n grows the curve hugs the x axis for longer and then spikes up to 1 at x equals 1. A big dot sits at the point 1 comma 1 where all the curves meet.xy11n=1n=2n=5n=20
The functions $x^n$ on $[0, 1]$ for $n = 1, 2, 5, 20$. As $n$ grows, each curve hugs the $x$-axis for longer and then rushes upward right at the end to meet $(1, 1)$. The pointwise limit is $0$ everywhere on $[0, 1)$ and jumps to $1$ at $x = 1$.

This is exactly the kind of thing that real analysis was built to sort out. Something about the way f_n converges to f at x = 0.99 is "slower" than at x = 0.5 — you need larger and larger n to get within any fixed tolerance as you move closer to 1. At x = 0.99, you need n around 459 to make x^n smaller than 0.01. At x = 0.999, you need n around 4603. The closer to 1, the worse it gets. Pointwise convergence hides this unevenness because the definition lets you pick N separately at each point.

Uniform convergence: the stronger definition

The fix is to demand that a single N works for every x in the domain simultaneously, not one N per point.

Uniform convergence

A sequence of functions f_n converges uniformly to f on a set D if, for every \varepsilon > 0, there exists a natural number N (depending only on \varepsilon, not on x) such that

n \geq N \;\;\Longrightarrow\;\; |f_n(x) - f(x)| < \varepsilon \text{ for every } x \in D.

Equivalently:

\sup_{x \in D} |f_n(x) - f(x)| \to 0 \text{ as } n \to \infty.

The quantifier order has swapped. In pointwise convergence, "for every x" comes first and then "there exists N" — the N depends on x. In uniform convergence, "there exists N" comes first and then "for every x" — the same N has to work for all points at once.

The second formulation in the definition is often the most useful in practice: the supremum of the pointwise errors must go to zero. If you can find even one x where the error stays big no matter how large n gets, uniform convergence fails.

Go back to f_n(x) = x^n on [0, 1). (Leave out x = 1 for now so the limit function is simply 0.) The pointwise error at any fixed x < 1 is |x^n - 0| = x^n, which goes to zero. But the supremum over all x \in [0, 1) is

\sup_{x \in [0, 1)} x^n = 1

because you can make x^n arbitrarily close to 1 by taking x close to 1 (even though you cannot reach 1). The supremum does not go to zero as n \to \infty — it stays pinned at 1. So the convergence is not uniform on [0, 1).

On the other hand, if you restrict to a slightly smaller interval, say [0, 1/2], then

\sup_{x \in [0, 1/2]} x^n = (1/2)^n \to 0.

So on [0, 1/2] the convergence is uniform. The same sequence of functions converges uniformly on one interval and non-uniformly on another.

Why uniform convergence preserves continuity

The reason this distinction matters: uniform convergence preserves continuity, pointwise does not.

Theorem. If each f_n is continuous on an interval I and f_n \to f uniformly on I, then f is continuous on I.

Proof sketch. Fix a \in I. You need to show that for every \varepsilon > 0 there exists \delta > 0 such that |x - a| < \delta implies |f(x) - f(a)| < \varepsilon. Split the difference into three pieces:

|f(x) - f(a)| = |f(x) - f_n(x) + f_n(x) - f_n(a) + f_n(a) - f(a)|
\leq |f(x) - f_n(x)| + |f_n(x) - f_n(a)| + |f_n(a) - f(a)|.

This is the standard "three-epsilon" trick. Now control each piece:

  1. Choose n so that |f_n(x) - f(x)| < \varepsilon/3 for every x. Uniform convergence guarantees such an n exists. The same n works for every x, including the specific a.
  2. With this n fixed, use the continuity of f_n to pick \delta > 0 such that |x - a| < \delta implies |f_n(x) - f_n(a)| < \varepsilon/3.
  3. The third term |f_n(a) - f(a)| is also less than \varepsilon/3 by the same uniform bound.

Adding the three pieces gives |f(x) - f(a)| < \varepsilon. So f is continuous at a.

The whole proof hangs on Step 1 — being able to pick a single n that controls the error for every x. Pointwise convergence cannot guarantee that. It lets you pick an n that works at a, and a (different) n that works at x, but not one that works at both. The argument collapses, and with it any hope of transferring continuity from the f_n to f.

The x^n example is the embodiment of this failure: every f_n is continuous, but the pointwise limit is not, because the convergence is not uniform near x = 1.

A second look: the hat example

Here is a more dramatic example that hammers the point home. Define

f_n(x) = \begin{cases} n^2 x & \text{if } 0 \leq x \leq 1/n \\ 2n - n^2 x & \text{if } 1/n < x \leq 2/n \\ 0 & \text{if } 2/n < x \leq 1. \end{cases}

Each f_n is a triangular "hat" on [0, 1]. The hat is zero outside the interval [0, 2/n], rises linearly to a peak of height n at x = 1/n, then drops linearly back to zero at x = 2/n.

Example 1: pointwise versus uniform on the hat sequence

Step 1. Show that f_n converges pointwise to 0.

Pick any x \in [0, 1].

If x = 0: f_n(0) = 0 for every n, so \lim f_n(0) = 0.

If x > 0: eventually 2/n < x (as soon as n > 2/x), and from that point on f_n(x) = 0. So the sequence f_n(x) is eventually zero and therefore has limit 0.

Why: at each fixed positive x, the hat is a moving target that eventually leaves x behind. After it passes, the function value at x stays at 0 forever.

Step 2. Find the supremum of |f_n - 0|.

The peak of f_n is at x = 1/n and has value n. So

\sup_{x \in [0,1]} |f_n(x) - 0| = n.

Why: the maximum of a continuous function on a bounded interval is achieved somewhere — here, at the peak of the hat. The formula gives the height of the peak directly.

Step 3. Check whether the supremum goes to zero.

n \to \infty \text{ as } n \to \infty.

It goes to infinity, not zero. So the convergence is not uniform — in fact, the maximum error grows even though the pointwise errors go to zero.

Why: uniform convergence requires the supremum of the errors to vanish. Here the errors at each point vanish individually, but the worst-case error gets worse and worse because the peak gets taller.

Result: f_n \to 0 pointwise on [0, 1], but not uniformly.

The hat functions f_nThree triangular hat functions f_2, f_4, and f_8 on the interval from 0 to 1. Each hat rises linearly to a peak at 1 over n and then falls back to zero at 2 over n. The peaks have heights 2, 4, and 8 respectively, getting taller as n grows and the base narrows.xy1f₂f₄f₈
Three hat functions in the sequence: $f_2$, $f_4$, $f_8$. Each has twice the height and half the base width of the previous one. At every fixed $x > 0$, the hats eventually stop touching $x$ and the value drops to zero. But the peak height grows without bound, so the worst-case error gets worse, not better.

If you had written down only the pointwise limit (f = 0) and tried to integrate it, you would get \int_0^1 f = 0. But each hat f_n has area \int_0^1 f_n = 1 — the area of a triangle with base 2/n and height n. So

\int_0^1 f_n \to 1, \qquad \int_0^1 f = 0.

The limit of the integrals does not equal the integral of the limit. Pointwise convergence cannot be trusted to commute with integration. Uniform convergence, on the other hand, always can: if f_n \to f uniformly on [a, b] and every f_n is integrable, then \int_a^b f_n \to \int_a^b f. This theorem (another one you will prove in a real analysis course) is why uniform convergence is the right concept for preserving the basic operations of calculus.

Series of functions

Just as series of numbers are infinite sums \sum a_n, you can form infinite sums of functions:

\sum_{n=0}^{\infty} u_n(x) \;=\; u_0(x) + u_1(x) + u_2(x) + \cdots

This is a series of functions. The partial sums

s_N(x) = \sum_{n=0}^{N} u_n(x)

form a sequence of functions; the series converges (pointwise or uniformly) depending on how the sequence of partial sums converges.

A series of functions is just a sequence of functions in disguise, and everything from the previous sections applies directly.

Power series

The most important example of a series of functions — the one you will actually use over and over — is the power series.

Power series

A power series centred at x_0 is a series of the form

\sum_{n=0}^{\infty} a_n (x - x_0)^n \;=\; a_0 + a_1(x - x_0) + a_2(x - x_0)^2 + \cdots

where a_0, a_1, a_2, \ldots are real numbers called the coefficients. When x_0 = 0, the series is simply \sum a_n x^n.

A power series is an infinite polynomial. Each term is a constant times a power of (x - x_0). For any specific value of x, you can ask: does the numerical series \sum a_n (x - x_0)^n converge?

The answer is beautifully structured. For any power series centred at x_0, there is a non-negative number R (possibly R = 0 or R = \infty) called the radius of convergence with the following property:

So a power series "lives" on an interval centred at x_0 of half-width R. Inside the interval, it defines a function. Outside, it is nonsense.

Example 2: the geometric power series

Take the power series

\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots

Step 1. Find the partial sums explicitly.

Each partial sum is a finite geometric series:

s_N(x) = 1 + x + x^2 + \cdots + x^N = \frac{1 - x^{N+1}}{1 - x} \quad (x \neq 1).

Why: the finite geometric sum formula gives a closed form for each partial sum, which is much easier to analyse than the series itself.

Step 2. Take the limit as N \to \infty.

For |x| < 1, x^{N+1} \to 0, so

\lim_{N \to \infty} s_N(x) = \frac{1}{1 - x}.

For |x| \geq 1, x^{N+1} does not go to zero, so the limit does not exist as a finite number and the series diverges.

Why: the whole convergence story is dictated by whether x^{N+1} shrinks to zero — which happens precisely when |x| < 1.

Step 3. Identify the radius of convergence.

The series converges exactly when |x| < 1, so the radius of convergence is R = 1.

Why: the interval of convergence is (-1, 1), half-width 1, so R = 1. At the endpoints x = \pm 1 the series diverges as well.

Step 4. Write down the limit function.

\sum_{n=0}^{\infty} x^n = \frac{1}{1 - x} \quad \text{for } |x| < 1.

Why: inside its radius of convergence, the power series is the function 1/(1-x). Outside, the function is still defined but the series representation breaks.

Result: The geometric power series has radius of convergence 1 and sums to 1/(1 - x) on (-1, 1).

The function $1/(1-x)$ as a solid black curve. The dashed red curves are the partial sums $s_1, s_2, s_3$ of the geometric series. The solid red curve is $s_4$, closer still. Inside the interval $(-1, 1)$, the partial sums hug the function. Outside, they spin away wildly. The radius of convergence $R = 1$ is marked by the vertical boundary at $x = \pm 1$.

A theorem you can take on faith for now. On any closed sub-interval [-r, r] strictly inside the interval of convergence (so r < R), the power series converges uniformly to its sum. That means on such an interval, term-by-term differentiation and integration are legal, and the sum function inherits continuity from the individual terms.

So inside its radius of convergence — strictly inside, not at the edge — a power series is as well-behaved as anything in calculus. You can differentiate it term by term and get a new power series for the derivative. You can integrate it term by term. You can substitute it into other power series. The whole machinery of infinite polynomials just works.

This is the engine that powers the next article: Taylor and Maclaurin series, which represent familiar functions like e^x, \sin x, and \cos x as explicit power series.

Common confusions

Going deeper

You now have the core definitions and examples. The rest of this section makes a few connections to the larger theory and addresses a question that probably occurred to you while reading.

The Weierstrass M-test

Given a series of functions \sum u_n(x), how do you check uniform convergence without finding the limit explicitly? The easiest tool is the Weierstrass M-test:

Theorem. If there exists a sequence of positive constants M_n such that |u_n(x)| \leq M_n for every x \in D, and \sum M_n converges, then \sum u_n(x) converges uniformly on D.

The idea: replace each function by its maximum value. If the series of maxima converges, then the original series is trapped below it and must also converge — and the convergence is uniform because the trap is the same at every point.

For the geometric series \sum x^n on [-r, r] with r < 1: you have |x^n| \leq r^n, and \sum r^n = 1/(1-r) < \infty. So by the M-test, the series converges uniformly on [-r, r]. This is exactly the kind of argument you use to establish uniform convergence on compact sub-intervals of the interval of convergence.

Computing the radius of convergence

For a specific power series \sum a_n (x - x_0)^n, the radius of convergence can often be computed by the ratio test:

R = \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right|

(when this limit exists). You recognise a power series as convergent where \lim |a_{n+1}(x-x_0)^{n+1} / a_n(x-x_0)^n| < 1 — which simplifies to |x - x_0| < \lim |a_n / a_{n+1}|.

For \sum x^n: a_n = 1 for every n, so |a_n/a_{n+1}| = 1 and R = 1. For \sum x^n / n!: |a_n/a_{n+1}| = (n+1)!/n! = n+1 \to \infty, so R = \infty — this series converges for every real x. You will meet this series as the Taylor expansion of e^x in the next article.

When the ratio test limit does not exist, a more general formula — the Cauchy-Hadamard formula 1/R = \limsup |a_n|^{1/n} — always works.

Why power series are so well-behaved

Inside its radius of convergence, a power series has a remarkable property: it is analytic. That means not only is it continuous, differentiable, and integrable, but its derivative is itself a power series with the same radius of convergence, obtained by differentiating term by term:

\frac{d}{dx}\sum_{n=0}^{\infty} a_n (x - x_0)^n = \sum_{n=1}^{\infty} n \cdot a_n (x - x_0)^{n-1}.

You can repeat this: the derivative is analytic, its derivative is analytic, and so on forever. Power series are infinitely differentiable inside their radius of convergence, and every derivative is computed by the same term-by-term rule.

This is one reason power series are so useful. Once you have a power series representation of a function, you have access to all its derivatives as straightforward algebraic operations on the coefficients. The calculus of analytic functions is almost trivial compared to general calculus.

Where this leads next

You know what it means for a sequence or series of functions to converge, why uniform convergence is the "good" kind, and how power series work inside their radius of convergence. The next article applies this machinery to express familiar functions as explicit power series.