1. What is a power series
An expression of the form
$$ \sum_{n=0}^{\infty} c_n (x - a)^n = c_0 + c_1 (x - a) + c_2 (x - a)^2 + c_3 (x - a)^3 + \cdots $$
where the $c_n$ are real numbers (the coefficients) and $a$ is a real number (the centre). For each fixed value of $x$ it becomes an ordinary numerical series, which may or may not converge.
Two things to notice straight away. First, the variable $x$ shows up exactly the way it does in a polynomial — to non-negative integer powers, multiplied by constants — except there are infinitely many terms. A power series is, quite literally, an infinite polynomial. Second, the "centre" $a$ is the point where the series is at its happiest: every term $(x - a)^n$ vanishes there except the constant $c_0$, so the series trivially equals $c_0$ at $x = a$. The further $x$ wanders from $a$, the more aggressively the series has to fight to converge.
The whole game of this topic is the tension between two questions:
- For which $x$ does the series converge? (Convergence theory.)
- When it does converge, what function does it equal? (Representation theory.)
We'll attack the first question with the ratio test and the second with Taylor's formula.
2. The prototype: the geometric series
The simplest power series — the one every other power series is secretly imitating — is the geometric series, with all coefficients equal to $1$:
$$ 1 + x + x^2 + x^3 + x^4 + \cdots = \sum_{n=0}^{\infty} x^n. $$From the algebra of finite geometric sums you already know the partial sums:
$$ S_N(x) = 1 + x + x^2 + \cdots + x^N = \frac{1 - x^{N+1}}{1 - x}, \qquad x \neq 1. $$If $|x| < 1$, then $x^{N+1} \to 0$ as $N \to \infty$, and the partial sums converge to a clean limit:
$$ \boxed{\,\sum_{n=0}^{\infty} x^n = \frac{1}{1 - x}\,} \qquad \text{for } |x| < 1. $$If $|x| \geq 1$, the terms $x^n$ don't shrink to zero — they stay the same size or blow up — and the series diverges. So this power series converges on the open interval $(-1, 1)$ and nowhere else. The function it represents, $1/(1-x)$, has a vertical asymptote at $x = 1$; the series feels that asymptote from far away and refuses to converge anywhere near it.
Almost every convergence proof for power series boils down to comparing your series with a geometric series. The ratio test, in particular, is "is your series eventually no worse than $\sum r^n$ for some $r < 1$?" If yes, you converge. If no, you don't. Hold on to $\sum x^n$ — it's the yardstick.
3. Radius and interval of convergence
The geometric series converges on $(-1, 1)$. What does the convergence set of a general power series look like? Remarkably, it's always symmetric around the centre $a$ — never some weird scattered subset.
For any power series $\sum c_n (x - a)^n$, there is a number $R \in [0, \infty]$ — the radius of convergence — such that:
- The series converges absolutely when $|x - a| < R$.
- The series diverges when $|x - a| > R$.
- At the boundary $|x - a| = R$, anything can happen — you have to check the two endpoints by hand.
$R = 0$ means the series converges only at $x = a$ itself; $R = \infty$ means it converges for every real $x$.
The interval of convergence is the actual set of $x$ where the series converges. It's always the open disk $(a - R, a + R)$ plus, possibly, one or both endpoints — so it's one of $(a-R, a+R)$, $[a-R, a+R)$, $(a-R, a+R]$, or $[a-R, a+R]$.
The geometric series is the cleanest example: centre $a = 0$, radius $R = 1$, and both endpoints $x = \pm 1$ diverge — interval of convergence $(-1, 1)$. Other series give endpoints that converge on one side, the other, or both.
4. Finding $R$ with the ratio test
The single most useful tool for computing radii is the ratio test applied to the absolute values of consecutive terms. For a power series $\sum c_n (x - a)^n$, look at
$$ L(x) = \lim_{n \to \infty} \left| \frac{c_{n+1}(x - a)^{n+1}}{c_n (x - a)^n} \right| = |x - a| \cdot \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|. $$The ratio test says the series converges absolutely when $L(x) < 1$ and diverges when $L(x) > 1$. Writing $\rho = \lim |c_{n+1}/c_n|$ when that limit exists, the inequality $L(x) < 1$ becomes
$$ |x - a| < \frac{1}{\rho}, \quad\text{so}\quad R = \frac{1}{\rho}. $$(If $\rho = 0$ then $R = \infty$; if $\rho = \infty$ then $R = 0$. Both cases come up.)
A quick gallery
| Series | $c_n$ | $\lim |c_{n+1}/c_n|$ | $R$ | Where it converges |
|---|---|---|---|---|
| $\sum x^n$ | $1$ | $1$ | $1$ | $(-1, 1)$ |
| $\sum x^n / n!$ | $1/n!$ | $1/(n+1) \to 0$ | $\infty$ | $(-\infty, \infty)$ |
| $\sum n! \, x^n$ | $n!$ | $n+1 \to \infty$ | $0$ | $\{0\}$ only |
| $\sum x^n / 3^n$ | $1/3^n$ | $1/3$ | $3$ | $(-3, 3)$ |
| $\sum (-1)^{n+1} x^n / n$ | $(-1)^{n+1}/n$ | $n/(n+1) \to 1$ | $1$ | $(-1, 1]$ — check endpoints! |
The last entry, the $\ln(1+x)$ series, is the canonical "you must check the endpoints" example. At $x = 1$ the series becomes the alternating harmonic series $1 - \tfrac12 + \tfrac13 - \cdots$, which converges. At $x = -1$ it becomes $-(1 + \tfrac12 + \tfrac13 + \cdots)$, the negative of the harmonic series, which diverges. The interval is $(-1, 1]$, not $(-1, 1)$.
The ratio test, by itself, tells you nothing about $|x - a| = R$ — it returns $L = 1$ there, which is the "inconclusive" case. You must analyse $x = a - R$ and $x = a + R$ as two separate ordinary numerical series, using whichever test fits (alternating series test, comparison, $p$-series, etc.).
5. Taylor and Maclaurin series
The convergence machinery is half the picture. The other half: which function does a power series represent? Equivalently — given a function $f$, can we find coefficients $c_n$ so that $f(x) = \sum c_n (x - a)^n$ near $x = a$?
If such an expansion exists, the coefficients are forced. Differentiate both sides $k$ times, set $x = a$, and only the $n = k$ term survives:
$$ f^{(k)}(a) = k! \, c_k \qquad\Longrightarrow\qquad c_k = \frac{f^{(k)}(a)}{k!}. $$That's the formula.
If $f$ has derivatives of every order at $a$, its Taylor series is
$$ T_a f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n. $$
When $a = 0$, the series is also called the Maclaurin series:
$$ T_0 f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n. $$
The Taylor series is, by construction, the unique power series whose value, slope, curvature, and every higher derivative at $a$ all agree exactly with $f$. It's the series that matches $f$ as well as it possibly can at the centre.
A subtle point: the Taylor series might exist, converge, and yet not equal $f$ away from $a$. A function that does equal its Taylor series in a neighbourhood of $a$ is called analytic at $a$. All the elementary functions you've met — polynomials, $e^x$, $\sin$, $\cos$, $\ln$, rational functions away from poles — are analytic on their natural domains. (The standard counterexample, $f(x) = e^{-1/x^2}$ with $f(0) = 0$, is smooth at $0$ but its Taylor series there is identically zero. Smooth $\neq$ analytic.)
Truncations and the error term
The degree-$n$ truncation,
$$ T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!} (x - a)^k, $$is the Taylor polynomial of degree $n$. It's an honest polynomial — a finite, computable approximation to $f$ near $a$. The remainder $R_n(x) = f(x) - T_n(x)$ measures how badly the truncation lies. Taylor's theorem with Lagrange remainder gives the bound:
$$ |R_n(x)| \leq \frac{M}{(n+1)!} |x - a|^{n+1}, \qquad M = \max_{[a, x]} \left| f^{(n+1)} \right|. $$That factorial in the denominator is the whole reason any of this works in practice. As $n$ grows, $(n+1)!$ obliterates $|x - a|^{n+1}$ for any bounded $x$ — and the Taylor polynomial converges to $f$ pointwise, often rapidly.
6. The five series to memorise
These are the bread and butter. You'll see them again in differential equations, complex analysis, Fourier analysis, numerical methods, statistical mechanics — everywhere. The derivations are all the same recipe: compute $f^{(n)}(0)$, divide by $n!$, recognise the pattern. The interesting thing is the patterns themselves.
$e^x$ — the fixed point of differentiation
Because $\frac{d}{dx} e^x = e^x$, every derivative at $0$ equals $e^0 = 1$. So $c_n = 1/n!$, and
$$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots $$The ratio of consecutive coefficients is $1/(n+1) \to 0$, so $R = \infty$ — this converges for every real (or, once we're brave enough, complex) $x$.
$\sin x$ and $\cos x$ — odd powers and even powers
The derivatives of sine cycle with period four: $\sin, \cos, -\sin, -\cos, \sin, \ldots$ At $x = 0$ they evaluate to $0, 1, 0, -1, 0, 1, \ldots$ — non-zero only at odd indices, with alternating signs. So all the even coefficients vanish and we get
$$ \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots $$Cosine works the same way, shifted by one: the odd derivatives vanish at $0$, leaving only even powers:
$$ \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots $$Both have $R = \infty$. (And you can check directly: term-by-term differentiation of the $\sin x$ series produces the $\cos x$ series, and differentiating $\cos x$ produces $-\sin x$. The familiar derivative rules are baked into the coefficients.)
$\ln(1 + x)$ — derived by integrating the geometric series
Here's the slick way. Start from the geometric series with $x$ replaced by $-x$:
$$ \frac{1}{1 + x} = 1 - x + x^2 - x^3 + x^4 - \cdots \qquad \text{for } |x| < 1. $$Now integrate both sides term by term from $0$ to $x$:
$$ \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n. $$Radius of convergence $R = 1$. Endpoints: converges at $x = 1$ (alternating harmonic series), diverges at $x = -1$ (negative harmonic series), so the interval of convergence is $(-1, 1]$.
$(1 + x)^k$ — the binomial series
For any real (or complex) exponent $k$, the function $f(x) = (1 + x)^k$ has derivatives $f^{(n)}(0) = k(k-1)(k-2)\cdots(k - n + 1)$. Defining the generalised binomial coefficient
$$ \binom{k}{n} = \frac{k(k-1)(k-2)\cdots(k - n + 1)}{n!}, $$the Maclaurin series is
$$ (1 + x)^k = \sum_{n=0}^{\infty} \binom{k}{n} x^n = 1 + kx + \frac{k(k-1)}{2!} x^2 + \frac{k(k-1)(k-2)}{3!} x^3 + \cdots $$This is the binomial series, valid for $|x| < 1$ (and at the endpoints in cases that depend on $k$). When $k$ is a non-negative integer the coefficients eventually become zero and the series collapses to the familiar finite binomial expansion. When $k = 1/2$ you get the Taylor series for $\sqrt{1+x}$; when $k = -1$ you recover the geometric series; when $k = -1/2$ you get the series Newton used to compute $\pi$.
7. Polynomials approximating $\sin x$
Memorising the $\sin x$ series is one thing. Seeing what its truncations do is another. Here are the Taylor polynomials of $\sin x$ at $a = 0$ of degree $1$, $3$, $5$, and $7$ drawn against $\sin x$ itself. Each successive polynomial picks up one more odd-power term, and each one tracks $\sin x$ accurately over a noticeably wider interval before peeling off.
Each new odd-power term keeps the Taylor polynomial glued to $\sin x$ for one more half-period of the sine wave. $T_1$ is correct only near zero; $T_7$ stays within a hair of $\sin x$ across $(-\pi, \pi)$ and beyond.
Two patterns are worth noticing. First, every polynomial agrees with $\sin x$ perfectly at $x = 0$ — they all pass through the origin with slope $1$, by construction. Second, each polynomial peels away from $\sin x$ further out, and the higher-degree ones peel away later. The window of usefulness grows with the degree. Taking enough terms, you can match $\sin x$ across any bounded interval to any required accuracy. That's $R = \infty$ in action.
8. Payoff: Euler's identity
Now the trick the complex-numbers page promised would land here. We have three series, all with $R = \infty$:
$$ \begin{aligned} e^x &= 1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \tfrac{x^4}{4!} + \tfrac{x^5}{5!} + \cdots \\ \cos x &= 1 - \tfrac{x^2}{2!} + \tfrac{x^4}{4!} - \tfrac{x^6}{6!} + \cdots \\ \sin x &= x - \tfrac{x^3}{3!} + \tfrac{x^5}{5!} - \tfrac{x^7}{7!} + \cdots \end{aligned} $$The cosine series uses every even power, with signs $+ - + - \cdots$. The sine series uses every odd power, with the same alternating pattern. The exponential uses every power, all with $+$ signs. The shapes are suspiciously compatible.
Substitute $x = i\theta$ into the $e^x$ series, and group terms by parity of the exponent. Use $i^0 = 1$, $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, repeating:
$$ \begin{aligned} e^{i\theta} &= 1 + i\theta - \tfrac{\theta^2}{2!} - i\tfrac{\theta^3}{3!} + \tfrac{\theta^4}{4!} + i\tfrac{\theta^5}{5!} - \cdots \\ &= \underbrace{\left(1 - \tfrac{\theta^2}{2!} + \tfrac{\theta^4}{4!} - \cdots\right)}_{\cos\theta} + i\underbrace{\left(\theta - \tfrac{\theta^3}{3!} + \tfrac{\theta^5}{5!} - \cdots\right)}_{\sin\theta} \\ &= \cos\theta + i\sin\theta. \end{aligned} $$That's Euler's formula: $e^{i\theta} = \cos\theta + i\sin\theta$. The series themselves did the work — no geometry required, no additional axioms, just the bookkeeping of even and odd powers of $i$.
Plug in $\theta = \pi$. Then $\cos\pi = -1$ and $\sin\pi = 0$, so
$$ e^{i\pi} = -1 + 0i = -1, \qquad\text{equivalently}\qquad \boxed{\,e^{i\pi} + 1 = 0.\,} $$The most famous equation in mathematics: $0$, $1$, $i$, $\pi$, $e$, addition, multiplication, exponentiation, and equality, all in seven symbols. The complex-numbers page took it as a "well-earned definition" and promised we'd justify it once power series were on the table. This is that justification.
We swept one thing under the rug: we plugged a complex number into a power series defined for real arguments. That step is legal — power series converge on disks in the complex plane just as they do on intervals on the real line, and $\sum x^n / n!$ has infinite radius in either world — but proving so belongs to complex analysis. For now, marvel: the algebra works out, and the answer is exactly the one the geometry of complex multiplication demanded.
9. Term-by-term differentiation and integration
Power series behave better than they have any right to. Inside the open interval of convergence $|x - a| < R$, you can differentiate and integrate a power series term by term, treating the infinite sum like a finite polynomial. The resulting series has the same radius $R$.
If $f(x) = \sum_{n=0}^{\infty} c_n (x - a)^n$ for $|x - a| < R$, then on the same open interval:
$$ f'(x) = \sum_{n=1}^{\infty} n\, c_n (x - a)^{n-1}, $$
$$ \int f(x)\, dx = C + \sum_{n=0}^{\infty} \frac{c_n}{n+1} (x - a)^{n+1}. $$
Both new series also have radius of convergence $R$.
This is enormously useful in practice. Instead of computing every derivative of $f$ at $a$ from scratch to find a new Taylor series, you can take a series you already know and operate on it. Three quick illustrations:
1. The derivative of $e^x$. Differentiate $\sum x^n / n!$ term by term:
$$ \frac{d}{dx} \sum_{n=0}^{\infty} \frac{x^n}{n!} = \sum_{n=1}^{\infty} \frac{n\, x^{n-1}}{n!} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!} = \sum_{m=0}^{\infty} \frac{x^m}{m!} = e^x. $$The series is its own derivative — the defining property of $e^x$ falls right out.
2. The series for $\arctan x$. Start from $\frac{d}{dx} \arctan x = 1/(1 + x^2)$. Use the geometric series with $-x^2$ in place of $x$:
$$ \frac{1}{1 + x^2} = 1 - x^2 + x^4 - x^6 + \cdots \qquad \text{for } |x| < 1. $$Integrate term by term from $0$ to $x$ (and use $\arctan 0 = 0$):
$$ \arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1}. $$Plug in $x = 1$ and (by the alternating series test) the series converges there to $\arctan 1 = \pi/4$, giving the Leibniz formula $\pi/4 = 1 - 1/3 + 1/5 - 1/7 + \cdots$ — a glorious, slow way to compute $\pi$ from nothing but odd reciprocals.
3. The series for $\ln(1+x)$ from $1/(1+x)$. This is the derivation we used in §6. The strategy generalises: integrate a known series, get a new one for the antiderivative.
Term-by-term operations are legal in the open interval $(a - R, a + R)$. At the boundary $x = a \pm R$ things get delicate — the differentiated series may diverge there even if the original converged, and behaviour at endpoints is not preserved. If you need to operate at an endpoint, you're past the safety zone of this rule and into more careful theorems (Abel's theorem, in particular).
10. Common pitfalls
The ratio test gives you $R$, the open interval. Whether the series converges at $x = a \pm R$ is a separate question — solved by examining the two endpoints as ordinary numerical series. Skipping this step is the most common way to write an "interval of convergence" that's missing one or both endpoints, or includes endpoints that diverge.
$\sin x$ uses only odd powers; $\cos x$ uses only even powers. Both alternate signs $+ - + - \cdots$. Mixing them up — putting an even power in the $\sin$ expansion, or the wrong sign on $x^4/4!$ in $\cos$ — is the classic exam-day disaster. Anchor yourself on $\sin 0 = 0$ (so no constant term, the series starts at $x$) and $\cos 0 = 1$ (so the series starts at $1$).
The series is the full infinite sum; the polynomial is a finite truncation. The series may equal $f$ exactly (when $f$ is analytic); the polynomial only approximates, with an error governed by the remainder. Saying "the Taylor polynomial of degree $5$ for $\sin x$ is $\sin x$" is wrong — it's an approximation that happens to be very good near $0$ and degrades further out.
A function can have derivatives of every order at $a$, yet not equal its Taylor series in any neighbourhood of $a$. The canonical counterexample is $f(x) = e^{-1/x^2}$ for $x \neq 0$ and $f(0) = 0$: every derivative at $0$ is exactly zero, so the Taylor series at $0$ is the zero series, yet $f$ is not the zero function. The lesson: "all derivatives exist" is not enough — analyticity is a strictly stronger property.
Inside $|x - a| < R$, everything is fine. Outside, the series doesn't even converge, so "differentiating it term by term" is operating on something that doesn't exist. And exactly at $|x - a| = R$, the rules are delicate. When in doubt, stay strictly inside.
11. Worked examples
Five examples. Try each before opening the solution. The goal isn't to match the final number — it's to make the procedure feel automatic.
Example 1 · Radius of convergence of $\sum n!\, x^n$
$c_n = n!$. Apply the ratio test:
$$ \left| \frac{c_{n+1}}{c_n} \right| = \frac{(n+1)!}{n!} = n + 1 \longrightarrow \infty. $$So $\rho = \infty$ and $R = 1/\rho = 0$. The series converges only at $x = 0$ — the factorial growth of the coefficients overwhelms any non-zero value of $x$. A perfectly legal power series that's nonetheless useless as a function: it represents nothing on an open set.
Example 2 · Interval of convergence of $\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n \cdot 2^n}$
$c_n = 1 / (n \cdot 2^n)$. Ratio:
$$ \left| \frac{c_{n+1}}{c_n} \right| = \frac{n \cdot 2^n}{(n+1) \cdot 2^{n+1}} = \frac{1}{2} \cdot \frac{n}{n+1} \longrightarrow \frac{1}{2}. $$So $R = 1 / (1/2) = 2$. The open interval is $(-2, 2)$. Check endpoints:
At $x = 2$: the series becomes $\sum \frac{2^n}{n \cdot 2^n} = \sum \frac{1}{n}$ — the harmonic series, which diverges.
At $x = -2$: the series becomes $\sum \frac{(-2)^n}{n \cdot 2^n} = \sum \frac{(-1)^n}{n}$ — the alternating harmonic series, which converges.
Interval of convergence: $[-2, 2)$.
Example 3 · Maclaurin series of $f(x) = e^{-x^2}$
Don't compute derivatives — substitute $-x^2$ in for $x$ in the series for $e^x$:
$$ e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots $$Radius of convergence: $\infty$, because the original $e^x$ series did, and the substitution $u = -x^2$ doesn't change that.
Bonus: this is precisely the series that makes the integral $\int e^{-x^2}\, dx$ — which has no elementary antiderivative — computable as a fast-converging series. The error function $\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2}\, dt$ is defined this way.
Example 4 · Approximate $\sin(0.3)$ using $T_3$ and bound the error
$T_3(x) = x - x^3/6$, so
$$ T_3(0.3) = 0.3 - \frac{(0.3)^3}{6} = 0.3 - \frac{0.027}{6} = 0.3 - 0.0045 = 0.2955. $$For the error, use Lagrange's bound with $M = \max |f^{(4)}| = \max |\sin x| \leq 1$:
$$ |R_3(0.3)| \leq \frac{1}{4!} (0.3)^4 = \frac{0.0081}{24} \approx 0.000338. $$So $\sin(0.3)$ is in $[0.2955 - 0.00034, \; 0.2955 + 0.00034] = [0.29516, 0.29584]$. The true value is $\sin(0.3) \approx 0.295520$ — comfortably inside the bound, and just two cubic-and-linear terms got us four digits of accuracy.
Example 5 · Use term-by-term integration to get the series for $\arctan x$
Start from the geometric series with $-x^2$ in place of $x$:
$$ \frac{1}{1 + x^2} = \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n} = 1 - x^2 + x^4 - x^6 + \cdots, \qquad |x| < 1. $$Integrate both sides from $0$ to $x$, term by term:
$$ \arctan x = \int_0^x \frac{dt}{1 + t^2} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots $$The radius of convergence is preserved: $R = 1$. At $x = 1$ the alternating series test gives convergence, so the interval is $[-1, 1]$. Setting $x = 1$ yields the Leibniz formula $\pi/4 = 1 - 1/3 + 1/5 - 1/7 + \cdots$, a delightful (if glacially slow) way to compute $\pi$.