Module 29: Power series and endpoint testing

8 exercises, each with a complete solution. 1 figure.

On this page

A. Prerequisites and observable goals
B. The central idea: an infinite polynomial with a domain
C. The radius theorem and legal operations
D. Why one radius organizes convergence
E. A four-stage workflow
F. Worked examples
G. Independent practice
H. Complete solutions: Module 29
I. Retention and readiness

A. Prerequisites and observable goals#

Before

Geometric series, ratio/root tests, alternating series, and interval notation.

After

Find a radius and interval of convergence, test endpoints separately, and differentiate or integrate a power series only where justified.

B. The central idea: an infinite polynomial with a domain#

A power series is a function assembled from infinitely many polynomial terms:

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

The coefficients $c_n$ are fixed numbers, $a$ is the center, and $x$ is the variable input. At each chosen $x$ , the expression becomes an ordinary numerical series. Its convergence can change with $x$ , so a power series comes with a convergence domain, not just a formula.

A power series has a domain. The partial sums

1+x+\cdots+x^N

for

N=2

(blue) and

N=6

(orange) hug

1/(1-x)

(black) more closely inside

-1<x<1

(yellow band) as

N

grows, and peel away from it outside. The series represents the function only on its interval of convergence.

C. The radius theorem and legal operations#

There is a radius $R\in[0,\infty]$ such that the series converges absolutely for $|x-a|<R$ and diverges for $|x-a|>R$ . When $0<R<\infty$ , the endpoints $a-R,a+R$ require separate tests. Neither, one, or both may converge, absolutely or conditionally. The center always converges because all positive powers vanish there.

Inside the open convergence interval, termwise differentiation and integration are valid:

f'(x)=\sum_{n=1}^{\infty}n c_n(x-a)^{n-1},\qquad \int_a^x f(t)\dd t=\sum_{n=0}^{\infty}\frac{c_n}{n+1}(x-a)^{n+1}.

The differentiated and integrated series have the same radius of convergence, but may have different endpoint behavior. These statements are not automatic permission to differentiate at a boundary where the original numerical series happens to converge. [O2Power]

The geometric identity

\frac1{1-z}=\sum_{n=0}^{\infty}z^n,\qquad |z|<1,

is a starting point for building many power-series representations by substitution, differentiation, and integration. Always translate $|z|<1$ into the actual $x$ condition.

D. Why one radius organizes convergence#

Suppose a power series converges at a nonzero displacement $x_0-a$ . Its terms there are bounded, say $|c_n(x_0-a)^n|\le M$ . At any strictly smaller displacement,

|c_n(x-a)^n|\le M\left|\frac{x-a}{x_0-a}\right|^n,

a convergent geometric bound. Thus convergence at one distance implies absolute convergence at every smaller distance. This nested structure explains the radius. The boundary itself is not covered by a strictly smaller-ratio argument.

A similar geometric domination on any closed subinterval strictly inside the radius makes the convergence controlled enough to interchange limits with differentiation or integration. The full uniform-convergence proof is optional, but the location restriction is essential.

E. A four-stage workflow#

Identify the center. Apply the ratio or root test with $x$ treated as a parameter, and solve the strict inequality giving absolute convergence. Translate it into an open interval. Then substitute each finite endpoint into the original series and classify that numerical series separately. After differentiating or integrating, retain the radius but retest the new endpoints rather than copying the old brackets.

F. Worked examples#

A shifted center. For $\sum_{n=1}^{\infty}(x-2)^n/(n4^n)$ , the absolute ratio tends to $|x-2|/4$ , so $R=4$ . At $x=-2$ , the series is $\sum(-1)^n/n$ , which converges conditionally. At $x=6$ , it is harmonic and diverges. The interval is $[-2,6)$ ; writing only $R=4$ would omit endpoint information.

Building a logarithm. On $|x|<1$ , substitute $z=-x$ into the geometric series and integrate from $0$ to $x$ :

\frac1{1+x}=\sum_{n=0}^{\infty}(-1)^n x^n, \qquad \ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^n}{n}.

The series converges at $x=1$ as well. Its value there equals $\ln2$ , but extending the interior identity to that endpoint needs an endpoint argument (for example, a remainder estimate from the finite geometric identity), not just the observation of convergence. The stretch problem develops a related justified endpoint evaluation.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

M29.01FSolution

Find the radius and interval of convergence of $\sum_{n=0}^{\infty}(x-1)^n/3^n$ , and its sum inside that interval.

M29.02CSolution

Find the interval of convergence of $\sum_{n=1}^{\infty}x^n/n^2$ , classifying both endpoints.

M29.03TSolution

Find the radius and interval of convergence of $\sum_{n=1}^{\infty}n(x+2)^n/5^n$ .

M29.04CSolution

Find the radius of convergence of $\sum_{n=0}^{\infty}x^n/n!$ and $\sum_{n=0}^{\infty}n!x^n$ .

M29.05TSolution

Derive a power series for $1/(1-x)^2$ centered at zero, and state its interval of convergence.

M29.06CSolution

Derive a power series for $\arctan x$ on $|x|<1$ by integrating a geometric series.

M29.07TSolution

The series $\sum_{n=1}^{\infty}x^n/n^2$ converges on $[-1,1]$ . Determine the interval of convergence of its termwise derivative, and explain why copying the original endpoints would fail.

M29.08SSolution

Use the finite geometric identity for $1/(1+t^2)$ and an integrated remainder bound on $[0,1]$ to prove $\pi/4=1-1/3+1/5-1/7+\cdots$ .

Optional small hints

For M29.03, the endpoint terms fail to tend to zero. For M29.07, differentiate the exponent as well as the coefficient. For M29.08, the remainder after powers $t^0$ through $t^{2N}$ is $(-1)^{N+1}t^{2N+2}/(1+t^2)$ .

H. Complete solutions: Module 29#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

M29.01Show solution

Translate the geometric ratio into an input interval

The ratio is $r=(x-1)/3$ . Thus $|x-1|<3$ , giving $\boxed{R=3}$ and open interval $(-2,4)$ . At either endpoint, terms have magnitude one and the series diverges. Inside, $\sum r^n=1/(1-r)=\boxed{3/(4-x)}$ . The rational function exists at many other inputs, but this particular series does not represent it outside its interval.

Back to M29.01

M29.02Show solution

Absolute convergence can persist at both endpoints

The ratio limit is $|x|$ , so $R=1$ . At $x=1$ , the series is $\sum1/n^2$ , convergent; at $x=-1$ , its absolute series is the same convergent $p$ -series. Thus the interval is $\boxed{[-1,1]}$ , with absolute convergence at both endpoints. The radius alone did not determine the brackets.

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M29.03Show solution

A polynomial coefficient changes endpoint behavior

The absolute ratio tends to $|x+2|/5$ , giving $R=5$ and interior $(-7,3)$ . At $x=3$ , terms are $n$ ; at $x=-7$ , they are $n(-1)^n$ . Neither tends to zero. Therefore the interval is $\boxed{(-7,3)}$ .

Back to M29.03

M29.04Show solution

Factorials in opposite places give opposite radii

For $x^n/n!$ , the absolute ratio is $|x|/(n+1)\to0$ for every real $x$ , so $\boxed{R=\infty}$ . For $n!x^n$ and fixed $x\ne0$ , the ratio is $(n+1)|x|\to\infty$ , proving divergence. At $x=0$ , only the constant term remains. Thus the second radius is $\boxed0$ . An infinite formal polynomial need not define a function on any open neighborhood of its center.

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M29.05Show solution

Differentiate only inside the known radius

For $|x|<1$ , differentiate $\sum_{n=0}^{\infty}x^n=1/(1-x)$ term by term:

\frac1{(1-x)^2}=\sum_{n=1}^{\infty}n x^{n-1} =\sum_{k=0}^{\infty}(k+1)x^k.

The radius stays one. At both endpoints the terms do not approach zero, so the interval is $\boxed{(-1,1)}$ . Reindexing $k=n-1$ explains the changed coefficient.

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M29.06Show solution

Integrate from a base point with known value

Using $z=-t^2$ , $1/(1+t^2)=\sum_{n=0}^{\infty}(-1)^n t^{2n}$ for $|t|<1$ . Integrating from $0$ to $x$ gives

\boxed{\arctan x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}},\qquad |x|<1.

The constant is zero because both sides vanish at $x=0$ . Endpoint evaluation is a separate justification, supplied for $x=1$ in M29.08.

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M29.07Show solution

The radius survives, the endpoints can change

The derivative series is $\sum_{n=1}^{\infty}x^{n-1}/n$ . Its radius remains one. At $x=1$ , it becomes the divergent harmonic series. At $x=-1$ , it is $\sum(-1)^{n-1}/n$ , which converges conditionally. Hence its interval is $\boxed{[-1,1)}$ . Differentiating a convergent numerical series at an endpoint is not justified merely by the original endpoint convergence.

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M29.08Show solution

Control the endpoint remainder explicitly

The finite identity is

\frac1{1+t^2}=\sum_{n=0}^N(-1)^n t^{2n} +\frac{(-1)^{N+1}t^{2N+2}}{1+t^2}.

Integrate over $[0,1]$ . The left side is $\arctan1=\pi/4$ , and the finite sum integrates to $\sum_{n=0}^N(-1)^n/(2n+1)$ . The remaining integral has absolute value at most $\int_0^1t^{2N+2}\dd t=1/(2N+3)\to0$ . Therefore the finite sums converge to $\boxed{\pi/4}$ . This proves the value at the boundary instead of assuming an interior identity extends there without explanation.

Back to M29.08

I. Retention and readiness#

Retrieve M28.06 and M26.03. Explain why radius, interval, and endpoint classification are three related but different outputs. Advance when you can change a power series by differentiation or integration and correctly retest its endpoints.