Chapter 9 · Power Series

9.1 Power Series and Functions

Series whose terms contain $x$ — where they converge, how wide that interval is, and how the geometric series turns fractions into infinite polynomials.

Polynomials are the friendliest functions in mathematics: we can evaluate them with nothing but addition and multiplication, and we can differentiate and integrate them in our heads. In Chapter 8 we met a sum that looks like a polynomial with infinitely many terms, $$1+x+x^2+x^3+\cdots=\sum_{n=0}^{\infty}x^n,$$ and found that this geometric series converges to $\dfrac{1}{1-x}$ whenever $|x|\lt 1$. So the function $\dfrac{1}{1-x}$ is an infinite polynomial — at least for $x$ between $-1$ and $1$.

This section takes that observation seriously. We define power series, discover that the set of $x$ for which one converges is always an interval centred at a point $a$, learn to find that interval with the ratio test, and then use the geometric series to write many other functions as power series. These skills are the foundation for everything in the rest of the chapter.

What Is a Power Series?

📘 Definition 9.1 — Power series

A power series centred at $a$ (or a power series in $x-a$) is a series 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 $x$ is a variable and the numbers $c_0,c_1,c_2,\dots$ are constants called the coefficients of the series. When $a=0$ the series is simply $\sum c_nx^n$, a power series centred at $0$.
Convention: we take $(x-a)^0=1$ even when $x=a$, so the first term is always $c_0$.

For each fixed number $x$, a power series is an ordinary series of numbers, which may converge or diverge. Where it converges, its sum defines a function $$f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n,$$ whose domain is the set of all $x$ for which the series converges. One value is always in that domain: at $x=a$ every term after the first is $0$, so the series converges to $c_0$. The interesting question is what happens away from the centre.

Some coefficients may be zero. For instance $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{4^n}=1-\frac{x^2}{4}+\frac{x^4}{16}-\cdots$ is a power series centred at $0$ with $c_1=c_3=c_5=\cdots=0$; and $\displaystyle\sum_{n=1}^{\infty}\frac{(2x-1)^n}{n^2}=\sum_{n=1}^{\infty}\frac{2^n}{n^2}\left(x-\tfrac12\right)^n$ is a power series centred at $a=\tfrac12$ — factor the coefficient of $x$ out of the bracket to see the centre.

Convergence of a Power Series

Suppose a power series centred at $a$ converges at some point $x=b\ne a$. Then its terms $c_n(b-a)^n$ tend to $0$, so they are bounded: $|c_n(b-a)^n|\le M$ for every $n$. For any $x$ closer to the centre, $|x-a|\lt|b-a|$, we get $$|c_n(x-a)^n|=|c_n(b-a)^n|\left|\frac{x-a}{b-a}\right|^n\le M\rho^n,\qquad \rho=\left|\frac{x-a}{b-a}\right|\lt 1,$$ so by comparison with a convergent geometric series the power series converges absolutely at $x$. Convergence at one point forces convergence at every point nearer the centre. Pushing this idea to its conclusion gives the central fact about power series.

📐 Theorem 9.1 — The three possibilities for convergence

For a power series $\displaystyle\sum_{n=0}^{\infty}c_n(x-a)^n$ exactly one of the following is true.

The series converges only at $x=a$.
The series converges absolutely for every real number $x$.
There is a real number $R\gt 0$ such that the series converges absolutely when $|x-a|\lt R$ and diverges when $|x-a|\gt R$. At the two points $x=a-R$ and $x=a+R$ the series may converge absolutely, converge conditionally, or diverge — the theorem says nothing about them.

📘 Definition 9.2 — Radius and interval of convergence

The number $R$ in case 3 of Theorem 9.1 is the radius of convergence; in case 1 we set $R=0$ and in case 2 we set $R=\infty$. The set of all $x$ at which the series converges is its interval of convergence. In case 3 it is one of the four intervals $$(a-R,\,a+R),\qquad [a-R,\,a+R),\qquad (a-R,\,a+R],\qquad [a-R,\,a+R];$$ in case 1 it is the single point $\{a\}$, and in case 2 it is $(-\infty,\infty)$.

🧭 Problem-solving strategy — finding the interval of convergence

Ratio test on the terms. With $a_n=c_n(x-a)^n$, compute $L(x)=\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$. Treat $x$ as a fixed number while taking the limit.
Solve $L(x)\lt 1$. Usually $L(x)=|x-a|/R$ for some $R$, which gives the radius. If $L(x)=0$ for every $x$ then $R=\infty$; if $L(x)=\infty$ for every $x\ne a$ then $R=0$.
Check each endpoint separately. At $x=a\pm R$ the ratio test gives $L=1$ and is silent. Substitute each endpoint into the series to get a series of numbers and test it with the divergence test, a $p$-series, comparison, or the alternating series test.
Write the interval, using square brackets only at endpoints where the series converges.

If every $c_n\ne0$ and $\displaystyle\lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|=\ell$ exists, step 1 gives $L(x)=\ell\,|x-a|$, so $R=1/\ell$ (with $R=\infty$ when $\ell=0$ and $R=0$ when $\ell=\infty$). When some coefficients are zero — for example a series with only even powers — use the ratio of consecutive non-zero terms instead.

✏️ Example 1 — The two extreme cases

Find the radius and interval of convergence of (a) $\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}$ and (b) $\displaystyle\sum_{n=0}^{\infty}n!\,x^n$.

💡 Show solution▶

(a) Ratio For fixed $x\ne0$, $$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{x^{n+1}}{(n+1)!}\cdot\frac{n!}{x^n}\right|=\frac{|x|}{n+1}\longrightarrow0\lt 1\quad(n\to\infty).$$

(a) Conclusion The limit is $0$ whatever $x$ is, so the series converges absolutely for every real $x$: $R=\infty$ and the interval of convergence is $(-\infty,\infty)$. (In 9.3 we will see that this series sums to $e^x$.)

(b) Ratio For fixed $x\ne0$, $$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(n+1)!\,x^{n+1}}{n!\,x^n}\right|=(n+1)|x|\longrightarrow\infty.$$

(b) Conclusion The series diverges for every $x\ne0$ and converges (to $c_0=1$) only at the centre: $R=0$, interval $\{0\}$. A power series like this one is useless for representing a function.

✏️ Example 2 — A half-open interval

Find the radius and interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(x-2)^n}{n\,3^n}$.

💡 Show solution▶

Ratio test $$\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x-2|^{n+1}}{(n+1)3^{n+1}}\cdot\frac{n\,3^n}{|x-2|^n}=\frac{|x-2|}{3}\cdot\frac{n}{n+1}\longrightarrow\frac{|x-2|}{3}.$$

Radius $\dfrac{|x-2|}{3}\lt 1\iff|x-2|\lt 3\iff -1\lt x\lt 5$. The centre is $a=2$ and $R=3$.

Right endpoint $x=5$: $\displaystyle\sum_{n=1}^{\infty}\frac{3^n}{n\,3^n}=\sum_{n=1}^{\infty}\frac1n$, the harmonic series, which diverges.

Left endpoint $x=-1$: $\displaystyle\sum_{n=1}^{\infty}\frac{(-3)^n}{n\,3^n}=\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$, the alternating harmonic series, which converges (conditionally) by the alternating series test.

Answer $R=3$; interval of convergence $[-1,\,5)$.

✏️ Example 3 — A closed interval, and finding the centre

Find the centre, radius and interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(2x-1)^n}{n^2}$.

💡 Show solution▶

Ratio test $$\left|\frac{a_{n+1}}{a_n}\right|=|2x-1|\cdot\frac{n^2}{(n+1)^2}\longrightarrow|2x-1|.$$

Endpoints $x=1$: $\displaystyle\sum\frac{1}{n^2}$, a convergent $p$-series ($p=2$). $x=0$: $\displaystyle\sum\frac{(-1)^n}{n^2}$, which converges absolutely because $\sum\frac1{n^2}$ does.

Answer Centre $\tfrac12$, $R=\tfrac12$, interval of convergence $[0,\,1]$.

⚠️ Common mistakes with intervals of convergence

1. Skipping the endpoints. The ratio test is always inconclusive at $|x-a|=R$. Examples 2 and 3 show that the endpoints really can behave differently — one in, one out, or both in.

2. Confusing $R$ with the length of the interval. $R$ is the distance from the centre to an endpoint; the interval has length $2R$.

3. Reading the radius off $|2x-1|\lt 1$ as $R=1$. Rewrite it as $|x-\tfrac12|\lt\tfrac12$ first.

4. Forgetting the centre always works. Even when $R=0$ the series converges at $x=a$.

Pick a series and slide the point $x$. The number line shows the interval of convergence (filled dot = endpoint included, open dot = excluded). The graph plots the partial sums $S_0,S_1,\dots,S_{40}$ at your $x$: inside the interval they settle to a horizontal level; outside they run away. Use the endpoint buttons to see the cases the ratio test cannot decide.

Series:$x$:

$x$—

ratio-test limit $L$—

$S_{40}$—

last term $|a_{40}|$—

Representing Functions as Power Series

Turning a function into a power series is useful because power series are so easy to compute with. Our master key is the geometric series.

📐 Theorem 9.2 — The geometric power series

$$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n=1+x+x^2+x^3+\cdots\qquad\text{for }|x|\lt 1,$$ and the series diverges for $|x|\ge1$. More precisely, for every $x\ne1$ and every $N\ge0$, $$\frac{1}{1-x}-\left(1+x+\cdots+x^N\right)=\frac{x^{N+1}}{1-x},$$ so when $|x|\lt 1$ the error after the term $x^N$ is exactly $\dfrac{x^{N+1}}{1-x}$, which tends to $0$.

The identity in Theorem 9.2 holds for every $x$ with $|x|\lt 1$, so we may replace $x$ by any expression $u$ and get $\dfrac{1}{1-u}=\sum u^n$ wherever $|u|\lt 1$. Together with multiplying by constants and powers of $x$ this produces a whole family of series.

🧭 Problem-solving strategy — using the geometric series

Make the shape $\dfrac{1}{1-u}$. Factor out constants so the denominator reads $1-(\text{something})$: e.g. $\dfrac{1}{2-x}=\dfrac12\cdot\dfrac{1}{1-x/2}$ and $\dfrac{1}{1+x^2}=\dfrac{1}{1-(-x^2)}$.
Substitute $u$ into $\sum u^n$ and simplify each term.
Multiply by any factor left over ($x^k$, a constant) term by term.
Interval: the series converges exactly when $|u|\lt 1$; solve that inequality for $x$. Multiplying by $x^k$ or a constant does not change it.

✏️ Example 4 — Substituting $-x^2$

Find a power series for $f(x)=\dfrac{1}{1+x^2}$ and its interval of convergence.

💡 Show solution▶

Shape $\dfrac{1}{1+x^2}=\dfrac{1}{1-\left(-x^2\right)}$, so $u=-x^2$.

Substitute $$\frac{1}{1+x^2}=\sum_{n=0}^{\infty}\left(-x^2\right)^n=\sum_{n=0}^{\infty}(-1)^nx^{2n}=1-x^2+x^4-x^6+\cdots$$

Interval Valid when $|-x^2|\lt 1$, i.e. $x^2\lt 1$, i.e. $-1\lt x\lt 1$. At $x=\pm1$ the series is $1-1+1-\cdots$, which diverges. Interval $(-1,1)$, $R=1$.

Notice $f(x)=\dfrac1{1+x^2}$ is perfectly well behaved for all $x$, yet its series centred at $0$ only works on $(-1,1)$. A power series can be more limited than the function it represents — Interactive 2 lets you watch this happen.

✏️ Example 5 — Factoring out a constant and multiplying by $x^3$

Find a power series for $f(x)=\dfrac{x^3}{2-x}$ and its interval of convergence.

💡 Show solution▶

Shape $$\frac{1}{2-x}=\frac{1}{2\left(1-\frac{x}{2}\right)}=\frac12\cdot\frac{1}{1-\frac x2}=\frac12\sum_{n=0}^{\infty}\left(\frac x2\right)^n=\sum_{n=0}^{\infty}\frac{x^n}{2^{n+1}},\qquad\left|\frac x2\right|\lt 1.$$

Multiply $$\frac{x^3}{2-x}=x^3\sum_{n=0}^{\infty}\frac{x^n}{2^{n+1}}=\sum_{n=0}^{\infty}\frac{x^{n+3}}{2^{n+1}}=\frac{x^3}{2}+\frac{x^4}{4}+\frac{x^5}{8}+\cdots$$

Interval $\left|\frac x2\right|\lt 1\iff|x|\lt 2$; at $x=\pm2$ the terms do not tend to $0$. Interval $(-2,2)$, $R=2$.

Check At $x=1$: $f(1)=1$, and $\tfrac12+\tfrac14+\tfrac18+\cdots=1$. ✓

✏️ Example 6 — A series centred at $a=1$

Write $f(x)=\dfrac1x$ as a power series centred at $a=1$, and give its interval of convergence.

💡 Show solution▶

Shape We need powers of $x-1$, so write $x=1+(x-1)$: $$\frac1x=\frac{1}{1+(x-1)}=\frac{1}{1-\left(-(x-1)\right)},\qquad u=-(x-1).$$

Substitute $$\frac1x=\sum_{n=0}^{\infty}(-1)^n(x-1)^n=1-(x-1)+(x-1)^2-(x-1)^3+\cdots$$

Interval Valid for $|x-1|\lt 1$, i.e. $0\lt x\lt 2$; at $x=0$ and $x=2$ the terms are $\pm1$ and the series diverges. Interval $(0,2)$, $R=1$ — the radius reaches exactly to the point $x=0$ where $f$ blows up.

Check At $x=1.5$: $1-0.5+0.25-0.125+\cdots=\dfrac{1}{1+0.5}=\dfrac23=\dfrac{1}{1.5}$. ✓

The dark curve is $f$; the blue curve is the partial sum $S_N$ (the first $N$ terms of its power series). The green band is the interval of convergence. Press Play to add terms: inside the band $S_N$ locks onto $f$; outside it the polynomial swings ever more wildly — even for $\tfrac{1}{1+x^2}$, whose graph is perfectly smooth there.

Function:terms $N$:6probe $x$:

probe $x$—

$f(x)$—

$S_N(x)$—

$|f(x)-S_N(x)|$—

Exercises

Find the radius of convergence and the interval of convergence. Check the endpoints separately.

1 $\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{3^n}$
2 $\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n}$
3 $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{\sqrt n}$
4 $\displaystyle\sum_{n=1}^{\infty}n\,x^n$
5 $\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n^2}$
6 $\displaystyle\sum_{n=0}^{\infty}n!\,(x-3)^n$
7 $\displaystyle\sum_{n=0}^{\infty}\frac{2^nx^n}{n!}$
8 $\displaystyle\sum_{n=1}^{\infty}\frac{(x-1)^n}{n\,2^n}$
9 $\displaystyle\sum_{n=1}^{\infty}\frac{(3x-2)^n}{n}$
10 $\displaystyle\sum_{n=1}^{\infty}\frac{n\,(x+2)^n}{5^n}$
11 $\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n}}{4^n}$
12 $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$
13 $\displaystyle\sum_{n=2}^{\infty}\frac{x^n}{n\ln n}$
14 $\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n^n}$ (hint: root test)

💡 Show solutions▶

1 Ratio $\to\dfrac{|x|}{3}$, so $R=3$. At $x=3$: $\sum1$; at $x=-3$: $\sum(-1)^n$ — both diverge (terms do not tend to $0$). Interval $(-3,3)$.

2 Ratio $=|x|\dfrac{n}{n+1}\to|x|$, $R=1$. $x=1$: harmonic series, diverges. $x=-1$: $\sum\frac{(-1)^n}{n}$ converges (AST). Interval $[-1,1)$.

3 Ratio $=|x|\sqrt{\dfrac{n}{n+1}}\to|x|$, $R=1$. $x=1$: $\sum\frac{(-1)^n}{\sqrt n}$ converges (AST). $x=-1$: $\sum\frac{1}{\sqrt n}$, a $p$-series with $p=\tfrac12\le1$, diverges. Interval $(-1,1]$.

4 Ratio $=|x|\dfrac{n+1}{n}\to|x|$, $R=1$. At $x=\pm1$ the terms $n(\pm1)^n$ do not tend to $0$: both diverge. Interval $(-1,1)$.

5 Ratio $\to|x|$, $R=1$. At $x=\pm1$, $\sum\frac{|\pm1|^n}{n^2}=\sum\frac1{n^2}$ converges, so both endpoints converge absolutely. Interval $[-1,1]$.

6 Ratio $=(n+1)|x-3|\to\infty$ for $x\ne3$. $R=0$; the series converges only at $x=3$.

7 Ratio $=\dfrac{2|x|}{n+1}\to0$ for every $x$. $R=\infty$, interval $(-\infty,\infty)$.

8 Ratio $\to\dfrac{|x-1|}{2}$, so centre $1$, $R=2$, open interval $(-1,3)$. $x=3$: $\sum\frac1n$ diverges; $x=-1$: $\sum\frac{(-1)^n}{n}$ converges. Interval $[-1,3)$.

9 Ratio $\to|3x-2|=3\left|x-\tfrac23\right|$. So $\left|x-\tfrac23\right|\lt\tfrac13$: centre $\tfrac23$, $R=\tfrac13$, open interval $\left(\tfrac13,1\right)$. $x=1$: $\sum\frac1n$ diverges; $x=\tfrac13$: $\sum\frac{(-1)^n}{n}$ converges. Interval $\left[\tfrac13,1\right)$.

10 Ratio $=\dfrac{n+1}{n}\cdot\dfrac{|x+2|}{5}\to\dfrac{|x+2|}{5}$: centre $-2$, $R=5$. At $x=3$ the terms are $n$; at $x=-7$ they are $(-1)^nn$ — neither tends to $0$. Interval $(-7,3)$.

11 Only even powers appear, so use consecutive terms: $\left|\dfrac{x^{2n+2}/4^{n+1}}{x^{2n}/4^n}\right|=\dfrac{x^2}{4}\lt 1\iff|x|\lt 2$. $R=2$. At $x=\pm2$: $\sum1$, diverges. Interval $(-2,2)$. (It is the geometric series $\sum(x^2/4)^n$.)

12 Consecutive terms: $\dfrac{x^2}{(2n+2)(2n+3)}\to0$ for every $x$. $R=\infty$, interval $(-\infty,\infty)$. (This is the series for $\sin x$.)

13 Ratio $=|x|\dfrac{n\ln n}{(n+1)\ln(n+1)}\to|x|$, $R=1$. $x=1$: $\sum\frac{1}{n\ln n}$ diverges by the integral test, since $\displaystyle\int_2^{\infty}\frac{dx}{x\ln x}=\lim_{b\to\infty}\left[\ln(\ln x)\right]_2^b=\infty$. $x=-1$: $\frac1{n\ln n}$ decreases to $0$, so the alternating series converges. Interval $[-1,1)$.

14 Root test: $\sqrt[n]{\left|\dfrac{x^n}{n^n}\right|}=\dfrac{|x|}{n}\to0\lt 1$ for every $x$. $R=\infty$, interval $(-\infty,\infty)$.

Use the geometric series to write each function as a power series (centred at $0$ unless stated). Give the interval of convergence.

1 $\dfrac{1}{1+x}$
2 $\dfrac{1}{1-x^2}$
3 $\dfrac{2}{3-x}$
4 $\dfrac{1}{1+4x^2}$
5 $\dfrac{x}{1-x}$
6 $\dfrac{x^2}{1+x^3}$
7 $\dfrac{1}{5+x}$
8 $\dfrac{3}{1+2x}$
9 $\dfrac{1}{x}$, centred at $a=2$
10 $\dfrac{x}{4+x^2}$
11 $\dfrac{1}{(1-x)(1-2x)}$ (use partial fractions first)

Check: inside the green band the partial sum of your series must lock onto the function

Part:terms $N$:8probe $x$:

probe $x$—

$f(x)$—

$S_N(x)$—

$|f(x)-S_N(x)|$—

💡 Show solutions▶

1 $u=-x$: $\dfrac{1}{1+x}=\displaystyle\sum_{n=0}^{\infty}(-1)^nx^n=1-x+x^2-\cdots$, valid for $|x|\lt 1$. Interval $(-1,1)$.

2 $u=x^2$: $\dfrac{1}{1-x^2}=\displaystyle\sum_{n=0}^{\infty}x^{2n}$, valid for $x^2\lt 1$. Interval $(-1,1)$.

3 $\dfrac{2}{3-x}=\dfrac23\cdot\dfrac{1}{1-x/3}=\displaystyle\sum_{n=0}^{\infty}\frac{2x^n}{3^{n+1}}$, valid for $|x/3|\lt 1$. Interval $(-3,3)$.

4 $u=-4x^2$: $\dfrac{1}{1+4x^2}=\displaystyle\sum_{n=0}^{\infty}(-1)^n4^nx^{2n}$, valid for $4x^2\lt 1$, i.e. $|x|\lt\tfrac12$. Interval $\left(-\tfrac12,\tfrac12\right)$.

5 $\dfrac{x}{1-x}=x\displaystyle\sum_{n=0}^{\infty}x^n=\sum_{n=0}^{\infty}x^{n+1}=\sum_{n=1}^{\infty}x^n$. Interval $(-1,1)$.

6 $u=-x^3$: $\dfrac{x^2}{1+x^3}=x^2\displaystyle\sum_{n=0}^{\infty}(-1)^nx^{3n}=\sum_{n=0}^{\infty}(-1)^nx^{3n+2}$, valid for $|x|^3\lt 1$. Interval $(-1,1)$.

7 $\dfrac{1}{5+x}=\dfrac15\cdot\dfrac{1}{1+x/5}=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{5^{n+1}}$, valid for $|x|\lt 5$. Interval $(-5,5)$.

8 $u=-2x$: $\dfrac{3}{1+2x}=\displaystyle\sum_{n=0}^{\infty}3(-2)^nx^n=\sum_{n=0}^{\infty}(-1)^n\,3\cdot2^nx^n$, valid for $|2x|\lt 1$. Interval $\left(-\tfrac12,\tfrac12\right)$.

9 $x=2+(x-2)$: $\dfrac1x=\dfrac12\cdot\dfrac{1}{1+\frac{x-2}{2}}=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n(x-2)^n}{2^{n+1}}$, valid for $|x-2|\lt 2$. Interval $(0,4)$.

10 $\dfrac{x}{4+x^2}=\dfrac x4\cdot\dfrac{1}{1+x^2/4}=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{4^{n+1}}$, valid for $x^2\lt 4$. Interval $(-2,2)$.

11 $\dfrac{1}{(1-x)(1-2x)}=\dfrac{2}{1-2x}-\dfrac{1}{1-x}$ (check: $2(1-x)-(1-2x)=1$). So the series is $\displaystyle\sum_{n=0}^{\infty}2\cdot2^nx^n-\sum_{n=0}^{\infty}x^n=\sum_{n=0}^{\infty}\left(2^{n+1}-1\right)x^n=1+3x+7x^2+15x^3+\cdots$. Both pieces converge only where $|2x|\lt 1$ and $|x|\lt 1$: interval $\left(-\tfrac12,\tfrac12\right)$.

III

Reasoning and applications.

1 A power series $\sum c_nx^n$ converges at $x=4$ and diverges at $x=-6$. What can you say about its behaviour at $x=-3$?
2 …at $x=7$?
3 …at $x=5$?
4 …at $x=-4$?
5 Give an example of a power series centred at $0$ whose interval of convergence is exactly $(-2,2]$.
6 Applied — repeated medication. A patient takes $100$ mg of a drug every $12$ hours, and a fraction $r$ ($0\lt r\lt 1$) of the drug in the body remains just before the next dose. The amount just after a very large number of doses is $A(r)=\sum_{n=0}^{\infty}100\,r^n$. Find a closed form for $A(r)$, say why the series converges, and evaluate it for $r=0.25$.
7 In part 6 with $r=0.25$, the amount just after the $k$-th dose is $100\left(1+r+\cdots+r^{k-1}\right)$. After how many doses is it within $1$ mg of the long-run value?
8 Applied — how many terms? A small processor approximates $\dfrac{1}{1-x}$ by $1+x+x^2+\cdots+x^N$. Using Theorem 9.2, find the smallest $N$ that makes the error less than $10^{-6}$ at $x=0.2$.
9 Show that if $\sum c_nx^n$ has radius of convergence $R$ (with $0\lt R\lt\infty$), then $\sum c_nx^{2n}$ has radius of convergence $\sqrt R$.

💡 Show solutions▶

1 Convergence at $x=4$ means $R\ge4$; divergence at $x=-6$ means $R\le6$. Since $|-3|\lt 4\le R$, the series converges absolutely at $x=-3$.

2 $|7|\gt 6\ge R$, so it diverges at $x=7$.

3 $4\le R\le6$ and $|5|$ lies in that range, so we cannot tell: it converges if $R\gt 5$, diverges if $R\lt 5$, and $x=5$ is an endpoint if $R=5$.

4 Cannot tell. If $R\gt 4$ the series converges at $-4$; but if $R=4$ then $-4$ is an endpoint, and convergence at one endpoint ($x=4$) says nothing about the other.

5 For example $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n\,2^n}$: the ratio test gives $R=2$; at $x=2$ it is $\sum\frac{(-1)^n}{n}$ (converges), at $x=-2$ it is $\sum\frac1n$ (diverges). Other answers are possible.

6 $A(r)=100\displaystyle\sum_{n=0}^{\infty}r^n=\frac{100}{1-r}$ mg, a geometric power series in $r$ that converges because $|r|\lt 1$. For $r=0.25$: $A=\dfrac{100}{0.75}\approx133.3$ mg.

7 By Theorem 9.2 the shortfall after $k$ doses is exactly $\dfrac{100\,r^k}{1-r}=133.3\times0.25^k$ mg. $k=3$: $2.08$ mg; $k=4$: $0.52$ mg $\lt 1$ mg. So after 4 doses.

8 Error $=\dfrac{0.2^{N+1}}{0.8}\lt 10^{-6}\iff0.2^{N+1}\lt 8\times10^{-7}\iff N+1\gt\dfrac{\ln(8\times10^{-7})}{\ln0.2}\approx8.72$. So $N+1=9$, $N=8$ (nine terms $1+x+\cdots+x^8$, error $6.4\times10^{-7}$; with $N=7$ the error is $3.2\times10^{-6}$).

9 Put $y=x^2$. The series $\sum c_ny^n$ converges absolutely for $|y|\lt R$ and diverges for $|y|\gt R$. So $\sum c_nx^{2n}$ converges absolutely when $x^2\lt R$, i.e. $|x|\lt\sqrt R$, and diverges when $|x|\gt\sqrt R$. Its radius is $\sqrt R$.

🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. When a patient takes the same tablet at regular intervals, how much drug builds up in the body, and how quickly is the “steady state” reached?

Why it matters. Hospital pharmacists and dosing guidelines around the world rely on this calculation to keep antibiotics, anti-epileptics and heart medicines above the effective level but below the toxic one.

What to do.

Model: a dose $D$ is given every $\tau$ hours and the drug has half-life $h$ hours, so a fraction $r=\left(\tfrac12\right)^{\tau/h}$ of the drug remains when the next dose is due. Explain why the amount just after the $(k+1)$-th dose is $D\left(1+r+r^2+\cdots+r^k\right)$.
Show that the long-run peak amount is the value of the power series $\sum_{n\ge0}Dr^n=\dfrac{D}{1-r}$, and justify convergence. Show that the long-run trough (just before a dose) is $\dfrac{Dr}{1-r}$.
Sample case: $D=500$ mg, $h=6$ h, $\tau=8$ h. Compute $r$, the steady peak and the steady trough.
Use Theorem 9.2 to find how many doses are needed before the peak reaches $95\%$ of its long-run value. Confirm with a spreadsheet that lists the first $10$ peaks.

Tools: Calculator, spreadsheet

Hand in: A half-page note with your formulas, the three numbers from step 3, the dose count from step 4, and one limitation of the model (for example, absorption is not instantaneous).

Going further. Find the dosing interval $\tau$ that keeps the steady trough above $250$ mg with $D=500$ mg, or explore a “loading dose” that reaches the steady state immediately.

Instructor note: $r=2^{-4/3}\approx0.397$; steady peak $\approx829$ mg, trough $\approx329$ mg; the peak first exceeds $95\%$ of $829$ mg at the 4th dose ($1-r^k\gt 0.95$ needs $k\ge4$; $k=3$ gives $93.75\%$).

🧪 Self-check quiz

0/5 correct

Q1. What are the centre and radius of convergence of $\displaystyle\sum_{n=0}^{\infty}\frac{(x-3)^n}{2^n}$?

Q2. Every power series $\sum c_n(x-a)^n$ is guaranteed to converge…

Q3. The ratio test applied to a power series gives the condition $\dfrac{|x+1|}{4}\lt 1$. Before the endpoints are checked, the series is known to converge on

Q4. Which is a power series for $\dfrac{1}{1+x}$, valid for $|x|\lt 1$?

Q5. The interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n}$ is

← Chapter 9 — Power Series 9.2 Properties of Power Series →