Series

Unit 3 Series

Key unit competence: Apply series in solving problems about approximations and limits of some mathematical expressions.

Key ideas

• A finite series ${\sum }_{k=1}^n{u}_k$. Useful sums: $\sum k=\frac{n(n+1)}{2}$, $\sum k^2=\frac{n(n+1)(2n+1)}{6}$, $\sum k^3={\left(\frac{n(n+1)}{2}\right)}^2$.
• An infinite series $\sum u_n$ converges if its partial sums $S_n$ have a finite limit $S$. If $\lim u_n\ne 0$, the series diverges.
• Geometric series ${\sum }_{n\ge 0}a{r}^n$ converges to $\frac{a}{1-r}$ if and only if $|r|<1$. The harmonic series $\sum \frac{1}{n}$ diverges; $\sum \frac{1}{n^p}$ converges iff $p>1$.
• Tests for positive series: comparison, ratio (d'Alembert): $L=\lim \left|\frac{u_{n+1}}{u_n}\right|$; $L<1$ converges, $L>1$ diverges, $L=1$ no conclusion.
• A power series $\sum a_n(x-c{)}^n$ converges for $|x-c|<R$ (radius of convergence); the end points must be checked separately.
• Taylor series at $x_0$: $f(x)=\sum \frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n$; at $x_0=0$ it is the Maclaurin series.
• Key Maclaurin series: $e^x=\sum \frac{x^n}{n!}$; $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots$; $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots$; $\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\dots$; $(1+x{)}^m=1+mx+\frac{m(m-1)}{2!}{x}^2+\dots$; $\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\dots$
• Uses: limits such as ${\lim }_{x\to 0}\frac{\sin x-x}{x^3}$, estimating $e$, $\pi$, $\ln 2$, $\sqrt{2}$, $\cos \frac{\pi }{6}$, and approximate roots of equations.

Formulas to know

Common misconceptions

• "If $u_n\to 0$ then $\sum u_n$ converges." No: $\sum \frac{1}{n}$ diverges although $\frac{1}{n}\to 0$.
• "The ratio test with $L=1$ proves divergence." No: it gives no conclusion.
• Mixing up the series of $\sin x$ (odd powers) and $\cos x$ (even powers).

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Find ${\sum }_{k=1}^{20}(3k+2)$.

   Show answer

   $3\times \frac{20\times 21}{2}+2\times 20=630+40=670$.

   Common mistake: Adding 2 only once instead of 20 times.

2. Use the series of $(1+x{)}^{1/3}$ to estimate $\sqrt[3]{1.03}$.

   Show answer

   $(1+x{)}^{1/3}\approx 1+\frac{x}{3}-\frac{x^2}{9}$; with $x=0.03$: $1+0.01-0.0001=1.0099$.

   Common mistake: Wrong coefficient $\frac{m(m-1)}{2}$: with $m=\frac{1}{3}$ it is $-\frac{1}{9}$.

3. Use the ratio test to study ${\sum }_{n\ge 1}\frac{n}{3^n}$.

   Show answer

   $\frac{u_{n+1}}{u_n}=\frac{n+1}{3n}\to \frac{1}{3}<1$: the series converges.

   Common mistake: Concluding divergence because $n$ grows.

4. Estimate $\ln 2$ using $\ln \frac{1+x}{1-x}=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots \right)$.

   Show answer

   $\frac{1+x}{1-x}=2$ gives $x=\frac{1}{3}$: $2\left(\frac{1}{3}+\frac{1}{81}+\frac{1}{1215}+\frac{1}{15309}+\cdots \right)\approx 0.6931$.

   Common mistake: Using $\ln (1+x)$ with $x=1$, which converges very slowly.

5. Write the Maclaurin series of $\cos x$ up to $x^4$ and estimate $\cos 0.2$.

   Show answer

   $\cos x\approx 1-\frac{x^2}{2}+\frac{x^4}{24}$; $\cos 0.2\approx 1-0.02+0.0000667=0.98007$.

   Common mistake: Using degrees instead of radians.

6. Estimate $e$ to 3 decimal places.

   Show answer

   $n=6$: $1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\frac{1}{720}\approx 2.71806$, so $e\approx 2.718$.

   Common mistake: Using too few terms.

7. Find the Taylor series of $e^x$ about $x=1$.

   Show answer

   All derivatives at 1 equal $e$: $e^x=e{\sum }_{n\ge 0}\frac{(x-1{)}^n}{n!}$.

   Common mistake: Writing powers of $x$ instead of $(x-1)$.

8. Evaluate ${\lim }_{x\to 0}\frac{e^x-1-x-\frac{x^2}{2}}{x^3}$.

   Show answer

   $e^x-1-x-\frac{x^2}{2}=\frac{x^3}{6}+\cdots$, so the limit is $\frac{1}{6}$.

   Common mistake: Stopping the series too early (at $x^2$).

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. The series ${\sum }_{n\ge 1}\frac{n}{n+1}$:	a. $3$
   2. Expand $(1-2x{)}^{-1/2}$ up to $x^2$.	b. $p>1$
   3. Find ${\sum }_{n=0}^{\infty }{\left(\frac{2}{3}\right)}^n$.	c. diverges, because the terms tend to 1
   4. For which $p$ does ${\sum }_{n\ge 1}\frac{1}{n^p}$ converge?	d. $1+x+\frac{3}{2}{x}^2$

   Show answer

   1 → c (diverges, because the terms tend to 1) · 2 → d ($1+x+\frac{3}{2}{x}^2$) · 3 → a ($3$) · 4 → b ($p>1$)

   • Step 1. $\lim \frac{n}{n+1}=1\ne 0$.
     Step 2. If the terms do not tend to 0, the series diverges ($n$-th term test).
   • Step 1. $(1+u{)}^{-1/2}=1-\frac{u}{2}+\frac{(-\frac{1}{2})(-\frac{3}{2})}{2}{u}^2=1-\frac{u}{2}+\frac{3}{8}{u}^2$.
     Step 2. $u=-2x$: $1+x+\frac{3}{8}(4x^2)=1+x+\frac{3}{2}{x}^2$.
   • Step 1. Geometric with $a=1$, $r=\frac{2}{3}$, $|r|<1$.
     Step 2. $S=\frac{a}{1-r}=\frac{1}{1/3}=3$.
   • Step 1. This is the $p$-series.
     Step 2. It converges exactly when $p>1$ ($p=1$ is the harmonic series, which diverges).

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $x-\frac{x^3}{3!}+\cdots$  ·  $x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$  ·  $1-x^2+x^4-x^6+\cdots$  ·  $e\left[1+(x-1)+\frac{(x-1{)}^2}{2}+\cdots \right]$
   1. The Taylor series of $e^x$ about $x=1$ begins .
   2. The Maclaurin series of $\frac{1}{1+x^2}$ is .
   3. The Maclaurin series of $\ln (1+x)$ begins .
   Show answer

   1: $e\left[1+(x-1)+\frac{(x-1{)}^2}{2}+\cdots \right]$ · 2: $1-x^2+x^4-x^6+\cdots$ · 3: $x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$

   • Step 1. All derivatives of $e^x$ at 1 equal $e$.
     Step 2. $\sum \frac{e}{n!}(x-1{)}^n=e\left[1+(x-1)+\frac{(x-1{)}^2}{2}+\cdots \right]$.
   • Step 1. $\frac{1}{1+u}=1-u+u^2-\cdots$ for $|u|<1$.
     Step 2. Replace $u$ by $x^2$.
   • Step 1. $f(0)=0$, $f^{\prime }(0)=1$, $f^{\prime \prime }(0)=-1$, $f^{\prime \prime \prime }(0)=2$.
     Step 2. $\frac{f^{(n)}(0)}{n!}$ gives $1,-\frac{1}{2},\frac{1}{3}$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Use three terms of the series of $\ln (1+x)$ to estimate $\ln 1.1$.	a. $0.0953$
   2. Find the coefficient of $x^3$ in the Maclaurin series of $\sin 2x$.	b. $\frac{3}{11}$
   3. Write $0.272727\dots$ as a fraction.	c. $\frac{1}{2}$
   4. Find ${\lim }_{x\to 0}\frac{e^x-1-x}{x^2}$.	d. $-\frac{4}{3}$

   Show answer

   1 → a ($0.0953$) · 2 → d ($-\frac{4}{3}$) · 3 → b ($\frac{3}{11}$) · 4 → c ($\frac{1}{2}$)

   • Step 1. $x=0.1$: $0.1-\frac{0.01}{2}+\frac{0.001}{3}$.
     Step 2. $0.1-0.005+0.000333=0.0953$.
   • Step 1. $\sin u=u-\frac{u^3}{6}+\cdots$ with $u=2x$.
     Step 2. $-\frac{(2x{)}^3}{6}=-\frac{8x^3}{6}=-\frac{4}{3}{x}^3$.
   • Step 1. $0.\overline{27}=\frac{27}{100}+\frac{27}{100^2}+\cdots =\frac{27/100}{1-1/100}$.
     Step 2. $=\frac{27}{99}=\frac{3}{11}$.
   • Step 1. $e^x-1-x=\frac{x^2}{2}+\frac{x^3}{6}+\cdots$.
     Step 2. Divide by $x^2$: $\frac{1}{2}+\frac{x}{6}+\cdots \to \frac{1}{2}$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $1-x^2+\frac{x^4}{2}-\cdots$  ·  6  ·  $1-x+\frac{x^2}{2}-\cdots$  ·  $[-1,1[$
   1. To estimate $e$ to 3 decimal places, the smallest $n$ with $\frac{1}{(n+1)!}<0.001$ is .
   2. The interval of convergence of ${\sum }_{n\ge 1}\frac{x^n}{n}$ is .
   3. The first terms of the Maclaurin series of $e^{-x^2}$ are .
   Show answer

   1: 6 · 2: $[-1,1[$ · 3: $1-x^2+\frac{x^4}{2}-\cdots$

   • Step 1. Need $(n+1)!>1000$.
     Step 2. $6!=720<1000$ but $7!=5040>1000$, so $n+1=7$ and $n=6$.
   • Step 1. Ratio: $\frac{n}{n+1}|x|\to |x|$, so $R=1$.
     Step 2. At $x=1$: $\sum \frac{1}{n}$ diverges. At $x=-1$: the alternating series converges. Interval $[-1,1[$.
   • Step 1. $e^u=1+u+\frac{u^2}{2}+\cdots$.
     Step 2. With $u=-x^2$: $1-x^2+\frac{x^4}{2}-\cdots$.

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