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The radius from the coefficients, the two ends tested separately, uniform convergence on compact pieces, and term-by-term calculus.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute the radius of convergence of a power series from its coefficients, test the two end points separately to give the full interval, and say where the convergence is uniform. You will also be able to justify differentiating and integrating a power series one term at a time inside its radius, which is the point at which every theorem of this course is spent at once.
The tests for series and the notion of absolute convergence; upper limits from unit 2; uniform convergence and what it preserves. This lesson is where all three are spent at once, and it is the last of the course.
A power series is written about a centre. Its radius of convergence is the distance from the centre within which it converges absolutely, and its interval of convergence is that interval together with whichever of the two end points converge.
A power series is $\sum_{n \ge 0} a_n (x - c)^n$. Write $$R = \frac{1}{\limsup_n |a_n|^{1/n}},$$ with the conventions $1/0 = \infty$ and $1/\infty = 0$. This is the radius of convergence, and the upper limit is used rather than a limit because the roots need not converge.
The trichotomy. The series converges absolutely for $|x - c| < R$, diverges for $|x - c| > R$, and at $|x - c| = R$ anything can happen. All three follow from the root test applied at a fixed $x$.
Uniform on compact pieces. For any $r < R$ the series converges uniformly on $|x - c| \le r$: the terms are bounded by $|a_n| r^n$, a convergent series of constants, and the Weierstrass test — itself the uniform Cauchy criterion — applies. It is usually not uniform on the whole open interval.
Term-by-term calculus. Inside the radius the sum $f$ is continuous, is differentiable with $f'(x) = \sum n a_n (x-c)^{n-1}$, and may be integrated term by term. The differentiated series has the same radius, since $n^{1/n} \to 1$ — which is what makes the derivative theorem's hypothesis available.
So a power series is as good as a polynomial strictly inside its radius, and the sum is infinitely differentiable there with $a_n = f^{(n)}(c)/n!$: every power series is the Taylor series of its own sum.
Another way: picture
Draw an interval of radius $R$ about the centre. Inside, the series behaves; outside, the terms grow without bound; and the two end points are two isolated questions the radius says nothing about. Shrink the interval by any amount and the convergence becomes uniform — which is why every theorem is stated on a closed piece strictly inside.
Another way: steps
To find where a power series converges:
All three of these have radius $1$ about $0$:
| Series | At $x = 1$ | At $x = -1$ | Interval |
|---|---|---|---|
| $\sum x^n$ | diverges | diverges | $(-1, 1)$ |
| $\sum x^n/n$ | diverges | converges | $[-1, 1)$ |
| $\sum x^n/n^2$ | converges | converges | $[-1, 1]$ |
Same radius, three different intervals. The ends are genuinely separate questions, settled by the tests of unit 2 — the $n$-th term test, the $p$-series comparison, the alternating series test — applied to ordinary numerical series.
Abel's theorem adds one fact worth knowing: if the series converges at an end point, its sum is continuous up to that end from inside. That is what makes $\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \cdots$ a legitimate evaluation of the logarithm's series at the edge of its interval, rather than a hopeful substitution.
The last sentence of the concept above rests on every unit of the course, and it is worth seeing the chain whole.
A first course in analysis is often described as making calculus rigorous. A better description is that it identifies the handful of statements — completeness, uniformity, and the exchange of two limits — that everything else in calculus was quietly assuming.
Claiming uniform convergence on the open interval. It holds on every closed piece strictly inside, and usually fails on the whole interval: $\sum x^n$ on $(-1,1)$ has partial sums whose distance from $1/(1-x)$ is unbounded.
Assuming the ends follow from the radius. They do not, and the three examples above have the same radius and three different intervals.
Assuming a smooth function equals its Taylor series. Every power series is the Taylor series of its sum, and not every smooth function is the sum of its own Taylor series. The flat function from the Taylor lesson is the counterexample.
Reading a small radius as a defect of the function. $1/(1+x^2)$ is smooth on all of $\mathbb{R}$ with radius $1$ about the origin. The obstruction is off the real line, at $\pm i$, and explaining that is the first business of complex analysis.
For $\sum \dfrac{x^n}{n 3^n}$ the roots of the coefficients tend to $1/3$, so the radius is $3$.
The radius, from the coefficients.
At $x = 3$ the series is $\sum 1/n$, which diverges.
One end, as a numerical series.
At $x = -3$ it is $\sum (-1)^n/n$, which converges. So the interval is $[-3, 3)$.
The other end, separately.
$\dfrac{1}{1-x} = \sum_{n\ge 0} x^n$ for $|x| < 1$, with radius $1$.
A known sum inside its radius.
Integrating term by term from $0$ to $x$, legitimate on any closed piece inside: $-\ln(1-x) = \sum_{n \ge 1} \dfrac{x^n}{n}$.
The integral exchanged with the sum.
Differentiating instead gives $\dfrac{1}{(1-x)^2} = \sum_{n \ge 1} n x^{n-1}$, with the same radius.
And the derivative, with the radius unchanged.
The roots of the coefficients tend to $1/4$, so the radius is $4$ about the centre $2$: convergence for $|x - 2| < 4$.
Radius first, about the stated centre.
At $x = 6$ the series is $\sum 1/n^2$, which converges; at $x = -2$ it is $\sum (-1)^n/n^2$, which converges absolutely.
Both ends, one at a time.
So the interval is the closed one from $-2$ to $6$. Here both ends came in, because the coefficients decay fast enough for absolute convergence at the edge — which is the case where the end points need no delicate argument at all.
Give the set of real $x$ for which $\displaystyle\sum_{n=1}^{\infty} \dfrac{x^n}{n\,8^{\,n}}$ converges.
This task has no paper form; do it on a device.
What is the radius of convergence of $\displaystyle\sum_{n=0}^{\infty} 9^{\,n} x^n$?
Answer:
Give the radius of convergence of each series.
| radius, or its reciprocal in the last row | |
|---|---|
| the radius for the coefficients raised to the power $n$ | |
| the radius for the reciprocals of those coefficients | |
| the reciprocal of the radius for the reciprocals of the factorials |
A power series has radius $4$. Match each region to what is known there.
| converges absolutely | converges uniformly | nothing general: it must be tested separately | diverges, since the terms do not tend to zero | |
|---|---|---|---|---|
| at a point strictly inside the radius | ||||
| on a closed interval strictly inside the radius | ||||
| at an end point of the interval | ||||
| beyond the radius |
Put the steps of finding the interval of convergence of a power series centred at $0$ with radius $7$ into order.
Number the steps in order (write the number in the box):
A power series has radius $3$. What licenses differentiating and integrating it one term at a time strictly inside that radius?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Give the set of real $x$ for which $\displaystyle\sum_{n=1}^{\infty} \dfrac{x^n}{n\,6^{\,n}}$ converges.
This task has no paper form; do it on a device.
You can find a radius, settle both ends, and say what licenses term-by-term calculus inside the radius. Say in your own words why the convergence is uniform on closed pieces inside but usually not on the whole interval. That is the end of the course: completeness, uniformity, and the exchange of two limits are what calculus was assuming all along.
10. Your turn: the interval of convergence of $\sum \dfrac{(x-2)^n}{n^2 4^n}$, step 3