Modules › Alternating series, power series, Taylor and Maclaurin series
Alternating series, power series, Taylor and Maclaurin series
Absolute and conditional convergence, radius and interval of convergence, Taylor series and error estimates.
Lectures 11–12 · Tutorial 12 · Sequences, series, power series and Taylor series
Textbook sections: Thomas ch. 10 Infinite Sequences and Series (10.1–10.10) · Stewart ch. 11 Infinite Sequences and Series (11.1–11.11)
Textbook sections: Thomas ch. 10 Infinite Sequences and Series (10.1–10.10) · Stewart ch. 11 Infinite Sequences and Series (11.1–11.11)
Alternating series, power series, Taylor and Maclaurin series
Calculus I · Module 13
After this module you can
- Decide absolute / conditional convergence of alternating series.
- Find radius and interval of convergence (check both endpoints).
- Build Taylor / Maclaurin series from known series and estimate the error.
Alternating series
- $\sum(-1)^nb_n$ converges if $b_n\downarrow0$ (AST). Error after $N$ terms: $|R_N|\le b_{N+1}$.
- Absolute: $\sum|a_n|$ converges (then $\sum a_n$ converges). Conditional: $\sum a_n$ converges but $\sum|a_n|$ diverges.
- $\sum(-1)^n\frac{\ln n}{n-\ln n}$ is conditionally convergent; $\sum\frac{(-2)^{n+1}}{n+5^n}$ is absolutely convergent.
- If $a_n\not\to0$ the series diverges: $\sum(-1)^{n+1}\left(\frac n{10}\right)^n$.
Radius and interval of convergence
- $\sum c_n(x-a)^n$: $R=\lim\left|\frac{c_n}{c_{n+1}}\right|$ (or $\frac1{\limsup\sqrt[n]{|c_n|}}$). Converges absolutely for $|x-a|<R$, diverges for $|x-a|>R$.
- Always test the endpoints $x=a\pm R$ separately.
- $\sum\frac{(x-1)^n}{\sqrt n}$: $R=1$; $x=2$ diverges, $x=0$ converges: interval $[0,2)$.
- $\sum n^nx^n$: $R=0$. $\sum\frac{x^n}{n!}$: $R=\infty$.
Taylor and Maclaurin series
| Function | Maclaurin series | Converges |
|---|---|---|
| $e^x$ | $\sum\frac{x^n}{n!}$ | all $x$ |
| $\sin x$ | $\sum\frac{(-1)^nx^{2n+1}}{(2n+1)!}$ | all $x$ |
| $\cos x$ | $\sum\frac{(-1)^nx^{2n}}{(2n)!}$ | all $x$ |
| $\frac1{1-x}$ | $\sum x^n$ | $|x|<1$ |
| $\ln(1+x)$ | $\sum\frac{(-1)^{n+1}x^n}{n}$ | $-1<x\le1$ |
Taylor at $a$: $c_n=\frac{f^{(n)}(a)}{n!}$. Substitute and differentiate/integrate known series term by term.
Building series and bounding the error
- $\frac{2x}{(1-x^2)^2}=\frac{d}{dx}\frac1{1-x^2}=\sum_{n\ge1}2nx^{2n-1}$. $\ln\frac{1+x}{1-x}=2\left(x+\frac{x^3}3+\frac{x^5}5+\cdots\right)$.
- $5\cos\pi x=\sum\frac{5(-1)^k(\pi x)^{2k}}{(2k)!}$; $2^x$ about $x=1$: $\sum\frac{2(\ln2)^n}{n!}(x-1)^n$.
- Lagrange: $|R_n(x)|\le\frac{M|x-a|^{n+1}}{(n+1)!}$ with $M\ge|f^{(n+1)}|$ between $a$ and $x$.
- $\ln1.1$ within $10^{-8}$: need $\frac{0.1^{n+1}}{n+1}<10^{-8}$, so $n=7$.
Typical mistakes
EndpointsForgetting to test $x=a\pm R$.
Conditional vs absoluteSay which one: AST alone gives conditional convergence at most.
Error boundsUse $M$ = max of the NEXT derivative on the interval.
Practice next
- Chapter quizzes Ch. 6 (3 sets)
- Practice bank, filtered to this chapter, then written mode
- Lab: Taylor polynomial and error
- Then the mock exams (Exams page)
Worked examples from the course problems
Worked example 1 · Alternating series (Tutorial 12, Q2(a))
$\displaystyle\sum_{n=1}^\infty\dfrac{(-2)^{n+1}}{n+5^n}$
Show the answer and the solution
Answer: converges absolutely
$\left|\frac{(-2)^{n+1}}{n+5^n}\right|\le2\left(\frac25\right)^n$, a convergent geometric series: comparison test ⇒ absolute convergence.
Worked example 2 · Power series (Tutorial 12, Q3(a))
$\displaystyle\sum_{n=1}^\infty\dfrac{(x-1)^n}{\sqrt n}$ has radius $R$ and interval of convergence
Show the answer and the solution
Answer: $R=1$, interval $[0,2)$
Ratio test: $|x-1|\lim\sqrt{\frac n{n+1}}=|x-1|<1$. At $x=2$: $\sum\frac1{\sqrt n}$ diverges; at $x=0$: $\sum\frac{(-1)^n}{\sqrt n}$ converges (AST). So $[0,2)$: absolutely for $0<x<2$, conditionally at $x=0$.
Worked example 3 · Taylor and Maclaurin series (Tutorial 12, Q4)
In the Taylor polynomial of $f(x)=\sin x$ about $a=\frac\pi4$, the coefficient of $\left(x-\frac\pi4\right)^2$ is
Show the answer and the solution
Answer: $-\dfrac{\sqrt2}{4}$
$c_2=\frac{f''(\pi/4)}{2!}=\frac{-\sin(\pi/4)}{2}=-\frac{\sqrt2}{4}$. The full cubic is $\frac{\sqrt2}{2}\left[1+u-\frac{u^2}2-\frac{u^3}6\right]$, $u=x-\frac\pi4$.
Worked example 4 · Alternating series (Tutorial 12, Q2(b))
$\displaystyle\sum_{n=1}^\infty(-1)^n\dfrac{\ln n}{n-\ln n}$
Show the answer and the solution
Answer: converges conditionally (not absolutely)
$b_n=\frac{\ln n}{n-\ln n}\to0$ and is eventually decreasing, so the series converges by the AST. But $\sum b_n$ diverges (compare with $\frac{1}{n}$: $b_n\ge\frac1n$ for $n\ge3$), so the convergence is conditional.
Practice
Quiz set A Quiz set B Quiz set C Practice bank (this chapter) Written mode Lab: Taylor polynomial and error