Calculus II built power series and Taylor series entirely out of real numbers. There was a variable x, and a center c on the real line. The radius of convergence was a distance along that same line. Chapter 8 of this course takes every one of those definitions and replaces x with z. Almost nothing else changes about how they work.
The one real surprise is saved for that chapter. What was an interval on a line becomes a whole patch of the plane. That happens because the variable is now free to move in two directions instead of one. Everything that finds the radius stays exactly the tool it always was.
A power series is an infinitely long polynomial. Its radius of convergence is a distance from the center. Inside that distance, the infinite sum still adds up to one ordinary finite number. Past that distance, it spirals off.
A Taylor series is a power series built for one specific job. It matches a function's value at one point. It also matches the slope there, the curvature, and every higher wiggle, all at once. An ordinary polynomial can only match finitely many of those at once.
Here is a gap worth knowing about. A Taylor series converging somewhere does not automatically mean it converges to the function that produced it. Those are two separate claims. A pathological function can satisfy the first while failing the second.
A power series centered at c is \displaystyle\sum_{n=0}^\infty a_n(x-c)^n. There is always some R\ge 0, possibly \infty, called the radius of convergence. The series converges absolutely for |x-c|<R. It diverges for |x-c|>R. The two endpoints x=c\pm R need separate testing, and they can go either way.
The standard way to find R is the ratio test applied to the series' own terms. It compares a_{n+1}(x-c)^{n+1} to a_n(x-c)^n. Then it solves for which x make the limit of that ratio less than 1.
A Taylor series of a function f at c is the specific power series \displaystyle\sum_{n=0}^\infty\frac{f^{(n)}(c)}{n!}(x-c)^n. It is built so that its coefficients match f's value and every derivative at c all at once. A Maclaurin series is the name for the case c=0.
Two facts must be kept apart. One is that a Taylor series converges. The other is that it converges to f(x) itself. The series can converge to a finite number that is not f(x). Every function this course meets from here on will turn out to have that gap close entirely.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
True or false — every power series converges at its own center x=c, regardless of the coefficients.
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Every term past the first carries a factor of (c-c)^n=0, no matter what the coefficients are — nothing can make that term nonzero.
No boundedness condition is needed — the (x-c)^n=0 factor for n\ge1 holds unconditionally at x=c.
This has nothing to do with whether the series happens to be geometric — it's true for every power series.
Why
At x=c, all terms past n=0 vanish identically, so convergence at the center is automatic and universal.
Plug x=c into a_n(x-c)^n and see what happens to every term past n=0.
From Calculus II — Integral Calculus & Infinite Series
Write the n=2 term (as a function of x only) of the power series \displaystyle\sum_{n=0}^\infty \frac{(x+3)^n}{n^2+1} centered at -3.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Substituting n=2 into the general term \frac{(x+3)^n}{n^2+1} gives \frac{(x+3)^2}{2^2+1}=\frac{(x+3)^2}{5}.
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Why
Substitute n=2 directly into the general term \dfrac{(x+3)^n}{n^2+1}.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
What distinguishes a Maclaurin series from a general Taylor series?
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Maclaurin series don't have a special convergence guarantee beyond ordinary Taylor series — that isn't the distinguishing feature.
Both Taylor and Maclaurin series are, in general, infinite — neither is automatically finite.
Maclaurin series can be built for any function with derivatives at 0, not only trig functions.
Why
A Maclaurin series is simply the special case of a Taylor series where the center c is chosen to be 0.
From Calculus II — Integral Calculus & Infinite Series
Find the coefficient of x^5 in the Maclaurin series for f(x)=e^{x}. Round to 6 decimal places.
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Why
Every derivative of e^x equals e^x, so f^{(n)}(0)=1 for all n, giving coefficient \frac{1}{n!}; for n=5, \frac{1}{120}\approx0.008333.
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Why
Every derivative of e^x equals e^x itself -- what does that make f^{(5)}(0), and what coefficient does that produce?
Radius of convergence, made visible
Drag the point along the axis. The partial sums stay settled inside the radius and blow up the moment you cross it.
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Worked example
Radius of convergence by the ratio test
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1
Find the radius of convergence of \displaystyle\sum_{n=1}^\infty \frac{(x-1)^n}{n\cdot 3^n}. \lim_{n\to\infty}\left|\frac{a_{n+1}(x-1)^{n+1}}{a_n(x-1)^n}\right| = \lim_{n\to\infty}\frac{n}{n+1}\cdot\frac13\cdot|x-1| = \frac{|x-1|}{3}
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The ratio test needs this limit below 1: \frac{|x-1|}{3} < 1 \quad\Longleftrightarrow\quad |x-1|<3
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3\boxed{R = 3, \text{ centered at } x=1}
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Worked example
A Maclaurin series with infinite radius
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1
Find the radius of convergence of the Maclaurin series for \cos x, \displaystyle\sum_{n=0}^\infty\frac{(-1)^n x^{2n}}{(2n)!}. \lim_{n\to\infty}\left|\frac{x^{2n+2}/(2n+2)!}{x^{2n}/(2n)!}\right| = \lim_{n\to\infty}\frac{x^2}{(2n+2)(2n+1)}
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The factorial in the denominator grows without bound no matter how large |x| is fixed to be, so this limit is 0 for every real x.
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A limit of 0 is below 1 regardless of x, so the series converges everywhere. \boxed{R=\infty}
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Where this goes next
Chapter 8's power series in \mathbb C, unchanged
Chapter 8 puts z in place of x in every definition above. Almost nothing changes about how R is found. It is the same ratio test, run against the same ratio of consecutive terms.
What does change is the shape. "Within distance R of the center" stops meaning an interval on a line. It starts meaning every point that close to the center, in any direction the plane allows. That is because z is now free to move in two directions instead of one.
Check your understanding (counts toward your progress)
What is the radius of convergence of the Maclaurin series for \sin x?
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R=1 would mean the series fails for, say, x=2 — but \sin(2) has a perfectly convergent series, just like every other real number.
\pi has no special role in where THIS series converges, even though \pi appears constantly in trig values.
R=0 would mean the series only converges at x=0 — but the ratio test's limit is 0 for every fixed x, not just x=0.
Why
The ratio test's limit is 0 for every real x. The factorial in the denominator grows without bound. So the series converges everywhere.
Run the ratio test on the series' own terms and see what the limit comes out to be, for a fixed x.
A student finds L=|x|/5<1 from the ratio test and concludes R=1/5. What's the mistake?
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The ratio test is exactly the standard tool for finding R — it applies here without issue.
L is just an intermediate expression in x before solving — R comes from finishing the inequality, not from L directly.
1/5 is the reciprocal of the correct answer. Solving |x|/5<1 correctly gives |x|<5.
Why
|x|/5<1 \Leftrightarrow |x|<5, so R=5. Reporting 1/5 mistakes the coefficient in the inequality for the radius itself.
Multiply both sides of |x|/5<1 by 5 before reading off R.
Why doesn't a Taylor series that converges somewhere automatically converge TO f(x) there?
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This is the exact claim that fails for certain pathological functions. The two facts have to be checked separately. They do coincide for every function this course discusses.
Taylor series do converge plenty of the time. That isn't the issue being asked about here.
Where the series is centered plays no role in this particular gap — it can occur at c=0 just as well.
Why
A series can converge to some finite number at a point without that number matching f(x). A separate argument, such as a remainder bound, is what closes the gap.
Ask what fact would have to be proven separately, beyond just 'the series converges somewhere.'
Find the radius of convergence of \displaystyle\sum \frac{(x-3)^n}{n^2\,4^n}, centered at 3.
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Why
\lim\dfrac{n^2}{(n+1)^2}\cdot\dfrac14|x-3|=\dfrac14|x-3|<1 \Leftrightarrow |x-3|<4.
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Why
The ratio test's n^2 factor doesn't affect the limit at all — only the base 4 matters.
Practice: Finding the radius, one series at a time (5 problems · doesn’t affect your progress)
The ratio test, applied to a power series' own terms, is used to find...
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The ratio test decides WHERE a series converges, not what value it converges to.
Whether a series is geometric is visible from its own shape directly — no test is needed to see that.
The center is simply read off the series' own notation, (x-c)^n — no test needed to find it.
Why
Solving the ratio test's inequality for x is exactly how the radius of convergence is found.
The radius of convergence
Determine the radius of convergence for the series \displaystyle\sum \frac{x^n}{5^n}.
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Why
\lim|x/5|<1 \Leftrightarrow |x|<5.
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Why
This series \displaystyle\sum \frac{(x+1)^n}{n\cdot 3^n} is centered at -1 — what is its radius of convergence?
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Why
\lim\dfrac13|x+1|<1 \Leftrightarrow |x+1|<3, so R=3.
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Why
The n in the denominator doesn't affect the ratio test's limit — only the base 3 does.
A power series has R=\infty. What does this mean?
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That describes the opposite case, R=0.
There are no endpoints to even worry about when R=\infty — the interval of convergence is the whole real line.
R=\infty is a completely ordinary outcome — e^x's own Maclaurin series is a standing example.
Why
R=\infty means the series converges for every real x, with nothing left to test at any endpoint.
It converges for every real x, with no exceptions
A student reports 'R=4, so the series converges for x in [-4,4]' without checking anything else. What's missing?
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Nothing in the setup suggests the radius was computed wrong — the issue is what happens AT the two boundary points.
This is actually a real, common outcome (like \sum x^n/n at x=-1) — it's just never guaranteed by the ratio test alone.
Endpoints sometimes converge and sometimes don't, depending on the specific series — that's exactly why they need individual testing.
Why
The ratio test only settles |x-c|<R versus |x-c|>R. Each endpoint, x=c\pm R, converges or diverges independently and must be tested on its own.
The endpoints x=\pm4 need SEPARATE testing — the ratio test says nothing about them