8.2 Infinite Series
Adding infinitely many terms by watching the partial sums $S_N$ โ geometric and telescoping series we can sum exactly, and the harmonic series that never stops growing.
What does $0.3333\ldots$ mean? It is shorthand for $\tfrac3{10}+\tfrac3{100}+\tfrac3{1000}+\cdots$, a sum with infinitely many terms โ and we all agree its value is $\tfrac13$. A ball that bounces forever travels a finite distance; a patient who takes a tablet every day forever carries a finite, steady amount of drug. Each of these is an infinite series, and each has a finite sum.
But not every infinite sum is finite: $1+1+1+\cdots$ obviously is not, and โ far less obviously โ neither is $1+\tfrac12+\tfrac13+\tfrac14+\cdots$. In this section we give "infinite sum" a precise meaning using the sequences of 8.1, sum the two families of series we can evaluate exactly (geometric and telescoping), prove that the harmonic series diverges, and learn the algebra of convergent series.
Partial sums and convergence
We cannot add infinitely many numbers one by one, but we can add the first $N$ of them and see what happens as $N$ grows. That turns a question about an infinite sum into a question about a sequence โ which we know how to answer.
๐ Definition 8.4 โ Infinite series, partial sums, convergence
Given a sequence $\{a_n\}$, the expression $$\sum_{n=1}^{\infty}a_n=a_1+a_2+a_3+\cdots$$ is an infinite series (or just a series); $a_n$ is its $n$th term. Its $N$th partial sum is the finite sum $$S_N=\sum_{n=1}^{N}a_n=a_1+a_2+\cdots+a_N.$$ If the sequence of partial sums converges, $S_N\to S$ (a real number), we say the series converges and has sum $S$, and write $\displaystyle\sum_{n=1}^{\infty}a_n=S$. If $\{S_N\}$ diverges, the series diverges; if $S_N\to\infty$ we write $\sum a_n=\infty$.
Two sequences are involved โ keep them apart. The terms $a_n$ are what we add; the partial sums $S_N$ are the running totals. Convergence of the series is about $S_N$, not about $a_n$. Series may start at any index (often $n=0$); the sum depends on where we start, but whether it converges does not.
โ๏ธ Example 1 โ Deciding convergence from the partial sums
Find a formula for $S_N$ and decide whether each series converges.
$$\text{(a) }\sum_{n=1}^{\infty}n\qquad\text{(b) }\sum_{n=1}^{\infty}(-1)^{n+1}\qquad\text{(c) }\sum_{n=1}^{\infty}\frac1{2^n}$$
๐ก Show solutionโถ
(a) $S_N=1+2+\cdots+N=\dfrac{N(N+1)}{2}\to\infty$. The series diverges.
(b) $S_1=1$, $S_2=1-1=0$, $S_3=1$, $S_4=0,\dots$ so $S_N=1$ for odd $N$ and $0$ for even $N$. The partial sums oscillate; the series diverges, even though they stay bounded.
(c) $S_1=\tfrac12$, $S_2=\tfrac34$, $S_3=\tfrac78$: the pattern is $S_N=1-\dfrac1{2^N}$ (the gap to $1$ halves each time, as in Zeno's walk). Since $\dfrac1{2^N}\to0$, $S_N\to1$: the series converges and $\displaystyle\sum_{n=1}^{\infty}\frac1{2^n}=1$.
Check $S_{10}=1-\tfrac1{1024}\approx0.99902$. โ
Geometric series
Example 1(c) is the simplest member of the most important family of series: each term is a fixed multiple $r$ of the one before.
๐ Theorem 8.8 โ Geometric series
Let $a\ne0$. The geometric series with first term $a$ and common ratio $r$, $$\sum_{n=1}^{\infty}ar^{\,n-1}=a+ar+ar^2+ar^3+\cdots,$$ converges if $|r|\lt1$, with sum $$\sum_{n=1}^{\infty}ar^{\,n-1}=\frac{a}{1-r}\qquad(|r|\lt1),$$ and diverges if $|r|\ge1$. Its partial sums are $S_N=a+ar+\cdots+ar^{N-1}=\dfrac{a\left(1-r^N\right)}{1-r}$ for $r\ne1$.
๐ก Show proofโถ
Shift and subtract $S_N=a+ar+\cdots+ar^{N-1}$ and $rS_N=ar+ar^2+\cdots+ar^{N}$. Subtracting, almost everything cancels: $S_N-rS_N=a-ar^N$, so for $r\ne1$, $S_N=\dfrac{a(1-r^N)}{1-r}$.
|r| < 1 $r^N\to0$ (Theorem 8.5), so $S_N\to\dfrac{a}{1-r}$.
|r| โฅ 1 If $r=1$, $S_N=Na\to\pm\infty$. If $r=-1$, $S_N$ alternates between $a$ and $0$. If $|r|\gt1$, $|r^N|\to\infty$, so $S_N$ is unbounded. In every case the series diverges.
๐งญ Summing a geometric series
- Check the ratio of consecutive terms is a constant $r$ (rewrite things like $\dfrac{2^{n+1}}{5^n}$ as $2\left(\tfrac25\right)^n$).
- Find the first term $a$ by substituting the starting index โ it is not always $n=1$.
- If $|r|\lt1$: $\ \text{sum}=\dfrac{\text{first term}}{1-\text{ratio}}$. If $|r|\ge1$: the series diverges.
โ๏ธ Example 2 โ Geometric series
Determine whether each series converges; if it does, find its sum.
$$\text{(a) }\sum_{n=0}^{\infty}5\left(-\frac13\right)^n\qquad\text{(b) }\sum_{n=1}^{\infty}\frac{2^{2n}}{3^{n+1}}\qquad\text{(c) }\sum_{n=1}^{\infty}\frac{2^{n+1}}{5^n}$$
๐ก Show solutionโถ
(a) First term ($n=0$): $a=5$. Ratio $r=-\tfrac13$, $|r|\lt1$. Sum $=\dfrac{5}{1-(-1/3)}=\dfrac{5}{4/3}=\dfrac{15}4$.
(b) Rewrite: $\dfrac{2^{2n}}{3^{n+1}}=\dfrac{4^n}{3\cdot3^n}=\dfrac13\left(\dfrac43\right)^n$. The ratio is $r=\tfrac43\gt1$, so the series diverges. (Its terms even grow.)
(c) $\dfrac{2^{n+1}}{5^n}=2\left(\dfrac25\right)^n$: ratio $r=\tfrac25$. First term ($n=1$): $a=\tfrac45$. Sum $=\dfrac{4/5}{1-2/5}=\dfrac{4/5}{3/5}=\dfrac43$.
Check (a) $S$ after 20 terms $\approx3.7500$; (c) after 20 terms $\approx1.33333$. โ
โ๏ธ Example 3 โ Repeating decimals are geometric series
Write as fractions: (a) $0.\overline{36}=0.363636\ldots$ (b) $2.4\overline{17}=2.4171717\ldots$
๐ก Show solutionโถ
(a) $0.\overline{36}=\dfrac{36}{100}+\dfrac{36}{100^2}+\dfrac{36}{100^3}+\cdots$, geometric with $a=\tfrac{36}{100}$, $r=\tfrac1{100}$: $$0.\overline{36}=\frac{36/100}{1-1/100}=\frac{36}{99}=\frac4{11}.$$
(b) Separate the non-repeating part: $2.4\overline{17}=2.4+0.0\overline{17}=\dfrac{24}{10}+\left(\dfrac{17}{1000}+\dfrac{17}{100\,000}+\cdots\right)$. The bracket has $a=\tfrac{17}{1000}$, $r=\tfrac1{100}$: $$\frac{17/1000}{99/100}=\frac{17}{990},\qquad 2.4\overline{17}=\frac{2376}{990}+\frac{17}{990}=\frac{2393}{990}.$$
Check $2393\div990=2.41717\ldots$ โ. The same reasoning gives $0.\overline{9}=\dfrac{9/10}{1-1/10}=1$: $0.999\ldots$ is $1$, not "just below" it.
Telescoping series
When each term is a difference $b_n-b_{n+k}$, the partial sum collapses like a folding telescope: almost every term cancels with a later one, leaving a short formula for $S_N$. Partial fractions (5.3) are the usual way to create the difference.
โ๏ธ Example 4 โ Telescoping series
(a) Find $\displaystyle\sum_{n=1}^{\infty}\frac{2}{n^2+4n+3}$. (b) Show that $\displaystyle\sum_{n=1}^{\infty}\ln\frac{n}{n+1}$ diverges.
๐ก Show solutionโถ
(a) Partial fractions $n^2+4n+3=(n+1)(n+3)$ and $\dfrac{2}{(n+1)(n+3)}=\dfrac{1}{n+1}-\dfrac{1}{n+3}$ (check: $\dfrac{(n+3)-(n+1)}{(n+1)(n+3)}=\dfrac2{(n+1)(n+3)}$ โ).
Collapse Write out $S_N$: $$S_N=\left(\tfrac12-\tfrac14\right)+\left(\tfrac13-\tfrac15\right)+\left(\tfrac14-\tfrac16\right)+\cdots+\left(\tfrac1N-\tfrac1{N+2}\right)+\left(\tfrac1{N+1}-\tfrac1{N+3}\right).$$ Each negative term cancels a positive term two brackets later. What survives is the first two positive and last two negative terms: $$S_N=\frac12+\frac13-\frac1{N+2}-\frac1{N+3}.$$
Limit $S_N\to\dfrac12+\dfrac13=\dfrac56$. So $\displaystyle\sum_{n=1}^{\infty}\frac2{n^2+4n+3}=\frac56$. (Check: $S_{1000}\approx0.83134$, and $\tfrac56\approx0.83333$; the gap $\tfrac1{1002}+\tfrac1{1003}\approx0.002$ โ.)
(b) $\ln\dfrac n{n+1}=\ln n-\ln(n+1)$, so $S_N=(\ln1-\ln2)+(\ln2-\ln3)+\cdots+(\ln N-\ln(N+1))=-\ln(N+1)\to-\infty$. The series diverges โ even though its terms $\ln\frac{n}{n+1}\to\ln1=0$.
โ ๏ธ Telescoping traps
Always write out enough terms at both ends of $S_N$ to see exactly what survives โ with a gap of two (as in Example 4) two terms survive at each end. And never rearrange the infinite series itself to "cancel" terms: work with the finite sum $S_N$, then take the limit.
The harmonic series diverges
In both Example 4(b) and the teaser of the chapter introduction the terms shrink to $0$ but the series still diverges. The most famous example is the harmonic series $\displaystyle\sum_{n=1}^{\infty}\frac1n$ (so called because the wavelengths of the overtones of a vibrating string are $1,\tfrac12,\tfrac13,\dots$ of the fundamental).
๐ Theorem 8.9 โ The harmonic series diverges
$$\sum_{n=1}^{\infty}\frac1n=1+\frac12+\frac13+\frac14+\cdots=\infty.$$ More precisely, $S_{2^k}\ge1+\dfrac k2$ for every $k\ge0$, so the partial sums are unbounded.
๐ก Show proof (grouping)โถ
Group Bracket the terms in blocks that end at powers of $2$: $$S_{16}=1+\tfrac12+\left(\tfrac13+\tfrac14\right)+\left(\tfrac15+\tfrac16+\tfrac17+\tfrac18\right)+\left(\tfrac19+\cdots+\tfrac1{16}\right).$$
Bound each block Every term in a block is at least the block's last term. So $\tfrac13+\tfrac14\ge\tfrac14+\tfrac14=\tfrac12$; $\ \tfrac15+\cdots+\tfrac18\ge4\cdot\tfrac18=\tfrac12$; $\ \tfrac19+\cdots+\tfrac1{16}\ge8\cdot\tfrac1{16}=\tfrac12$. Each block contributes at least $\tfrac12$.
Conclude $S_{2^k}\ge1+\tfrac12+\underbrace{\tfrac12+\cdots+\tfrac12}_{k-1\text{ blocks}}=1+\tfrac k2$. Since $1+\tfrac k2\to\infty$, the increasing sequence $S_N$ is unbounded and diverges to $\infty$ (Theorem 8.6). โ
How slowly? $S_N$ grows roughly like $\ln N+0.5772$. The first partial sum above $10$ is $S_{12\,367}$; to pass $100$ you would need more than $10^{43}$ terms.
โ ๏ธ The most common mistake in this chapter
"The terms go to $0$, so the series converges" is false. What is true (proved in 8.3) is the reverse: if $\sum a_n$ converges, then $a_n\to0$. So $a_n\to0$ is necessary for convergence but not sufficient โ the harmonic series is the standard counterexample.
The algebra of convergent series
Because a series is a limit of finite sums, the limit laws for sequences give us rules for combining series.
๐ Theorem 8.10 โ Properties of convergent series
Suppose $\displaystyle\sum_{n=1}^\infty a_n=A$ and $\displaystyle\sum_{n=1}^\infty b_n=B$ both converge, and $c$ is a constant. Then
$$\sum_{n=1}^{\infty}(a_n+b_n)=A+B,\qquad\sum_{n=1}^{\infty}(a_n-b_n)=A-B,\qquad\sum_{n=1}^{\infty}c\,a_n=cA.$$
Moreover:
- if $\sum a_n$ converges and $\sum b_n$ diverges, then $\sum(a_n+b_n)$ diverges;
- if $c\ne0$ and $\sum a_n$ diverges, then $\sum c\,a_n$ diverges;
- adding, removing or changing finitely many terms does not affect whether a series converges (although it changes the sum).
โ ๏ธ What the theorem does not say
No product rule: $\sum a_nb_n$ is generally not $AB$. With $a_n=b_n=\tfrac1{2^n}$: $A=B=1$, but $\sum\tfrac1{4^n}=\tfrac13\ne1$.
Two divergent series can add to a convergent one: $\sum1$ and $\sum(-1)$ both diverge, but $\sum(1-1)=0$. So "divergent + divergent" tells you nothing.
โ๏ธ Example 5 โ Combining series
$$\text{(a) }\sum_{n=1}^{\infty}\left(\frac3{n(n+1)}+\frac{2^n}{5^n}\right)\qquad\text{(b) }\sum_{n=1}^{\infty}\left(\frac1n-\frac1{2^n}\right)\qquad\text{(c) }\sum_{n=4}^{\infty}\frac1{n(n+1)}$$
๐ก Show solutionโถ
Fact $\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}$ telescopes: $S_N=1-\dfrac1{N+1}$, so $\displaystyle\sum_{n=1}^\infty\frac1{n(n+1)}=1$.
(a) Both pieces converge: $\displaystyle\sum\frac3{n(n+1)}=3(1)=3$ and $\displaystyle\sum_{n=1}^\infty\left(\tfrac25\right)^n=\frac{2/5}{1-2/5}=\frac23$. By Theorem 8.10 the sum is $3+\tfrac23=\dfrac{11}3$.
(b) $\sum\tfrac1{2^n}$ converges (to $1$) but $\sum\tfrac1n$ diverges, so their difference diverges (convergent + divergent).
(c) Removing the first three terms of a convergent series: $$\sum_{n=4}^{\infty}\frac1{n(n+1)}=1-\left(\frac12+\frac16+\frac1{12}\right)=1-\frac34=\frac14.$$ (Or telescope directly from $n=4$: only $\tfrac14$ survives.)
Applications: accumulation
Geometric series appear whenever a quantity is repeatedly scaled by the same factor and the pieces are added up: bounces of a ball, doses of a drug, rounds of spending in an economy (the "multiplier effect"), or payments discounted year after year.
โ๏ธ Example 6 โ A bouncing ball and a daily medicine
(a) A ball is dropped from a height of $10$ m. Each time it hits the ground it rebounds to $\tfrac34$ of the height it fell from. Assuming it bounces forever, find the total vertical distance it travels.
(b) A patient takes $100$ mg of a drug every 24 hours. After 24 hours, $25\%$ of the drug present remains in the body. Find the amount just after the $n$th dose, and the long-run amounts just after and just before a dose.
๐ก Show solutionโถ
(a) Model Down $10$; then up and down $7.5$, up and down $5.625$, โฆ Each rebound height is $\tfrac34$ of the previous: $$D=10+2\left(7.5+7.5\cdot\tfrac34+7.5\left(\tfrac34\right)^2+\cdots\right).$$
(a) Sum The bracket is geometric with $a=7.5$, $r=\tfrac34$: $\dfrac{7.5}{1-3/4}=30$. So $D=10+2(30)=70$ m.
(b) Model Just after the $n$th dose the body contains the new dose plus $25\%$ of the previous dose, $25\%$ of $25\%$ of the one before, and so on: $$Q_n=100+100(0.25)+100(0.25)^2+\cdots+100(0.25)^{n-1}=\frac{100\left(1-0.25^n\right)}{1-0.25}\ \text{mg}.$$ So $Q_1=100$, $Q_2=125$, $Q_3=131.25$ mg.
(b) Long run $Q_n\to\dfrac{100}{0.75}\approx133.3$ mg just after a dose; just before the next dose $25\%$ of this remains, about $33.3$ mg. The level is already within $0.2$ mg of the limit after $5$ doses ($Q_5\approx133.20$ mg). This is a recursion $Q_{n+1}=0.25Q_n+100$ of the kind met in 8.1 (Exercise IV.1), now summed explicitly.
Exercises
I
Geometric series. In 1โ8 determine whether the series converges, and if so find its sum.
- 1 $\displaystyle\sum_{n=1}^{\infty}\left(\frac13\right)^n$
- 2 $\displaystyle\sum_{n=0}^{\infty}4\left(-\frac12\right)^n$
- 3 $\displaystyle\sum_{n=1}^{\infty}\frac{3^n}{2^n}$
- 4 $\displaystyle\sum_{n=1}^{\infty}\frac{5}{10^n}$
- 5 $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n2^{n+1}}{7^n}$
- 6 $\displaystyle\sum_{n=2}^{\infty}e^{-n}$
- 7 $\displaystyle\sum_{n=1}^{\infty}\frac{\pi^n}{3^n}$
- 8 $1-0.2+0.04-0.008+\cdots$
- 9 For which $x$ does $\displaystyle\sum_{n=0}^{\infty}\frac{(x-1)^n}{2^n}$ converge? Find its sum as a function of $x$.
- 10 For $\displaystyle\sum_{n=1}^{\infty}2(0.9)^{n-1}$, find $S_N$, the sum $S$, and the smallest $N$ with $S-S_N\lt0.01$.
Numerical check: partial sums $S_N$ for large $N$
๐ก Show solutionsโถ
1 $a=\tfrac13$, $r=\tfrac13$: $\dfrac{1/3}{2/3}=\dfrac12$.
2 $a=4$, $r=-\tfrac12$: $\dfrac4{3/2}=\dfrac83$.
3 $r=\tfrac32\gt1$: diverges.
4 $a=\tfrac12$, $r=\tfrac1{10}$: $\dfrac{1/2}{9/10}=\dfrac59$ (that is, $0.555\ldots$).
5 $\dfrac{(-1)^n2^{n+1}}{7^n}=2\left(-\tfrac27\right)^n$; $a=2$ (at $n=0$), $r=-\tfrac27$: $\dfrac{2}{9/7}=\dfrac{14}9$.
6 $a=e^{-2}$, $r=e^{-1}$: $\dfrac{e^{-2}}{1-e^{-1}}=\dfrac1{e^2-e}\approx0.2141$.
7 $r=\tfrac\pi3\approx1.047\gt1$: diverges (only just!).
8 $a=1$, $r=-0.2$: $\dfrac1{1.2}=\dfrac56$.
9 Geometric with $a=1$, $r=\dfrac{x-1}2$. Converges iff $\left|\dfrac{x-1}2\right|\lt1\iff-1\lt x\lt3$, with sum $\dfrac1{1-\frac{x-1}2}=\dfrac{2}{3-x}$. (E.g. $x=2.2$: sum $2.5$.)
10 $S_N=\dfrac{2(1-0.9^N)}{0.1}=20\left(1-0.9^N\right)$, $S=20$, and $S-S_N=20(0.9)^N\lt0.01\iff0.9^N\lt0.0005\iff N\gt\dfrac{\ln0.0005}{\ln0.9}\approx72.14$. Smallest $N=73$ (check: $20(0.9)^{72}\approx0.0102$, $20(0.9)^{73}\approx0.0091$).
II
Repeating decimals. Write each as a fraction in lowest terms by summing a geometric series.
- 1 $0.\overline{7}$
- 2 $0.\overline{12}$
- 3 $1.\overline{234}$
- 4 $0.4\overline{5}$
- 5 Explain, using a geometric series, why $0.\overline{9}=1$.
๐ก Show solutionsโถ
1 $\dfrac{7/10}{1-1/10}=\dfrac79$.
2 $\dfrac{12/100}{1-1/100}=\dfrac{12}{99}=\dfrac4{33}$.
3 $1+\dfrac{234/1000}{1-1/1000}=1+\dfrac{234}{999}=1+\dfrac{26}{111}=\dfrac{137}{111}$ ($\approx1.234234$ โ).
4 $\dfrac4{10}+\dfrac{5/100}{1-1/10}=\dfrac4{10}+\dfrac5{90}=\dfrac{36+5}{90}=\dfrac{41}{90}$ ($\approx0.4556$ โ).
5 $0.\overline9=\sum_{n=1}^\infty\dfrac9{10^n}$, geometric with $a=\tfrac9{10}$, $r=\tfrac1{10}$: sum $\dfrac{9/10}{9/10}=1$. The partial sums $0.9,0.99,0.999,\dots$ equal $1-10^{-N}\to1$; the infinite decimal means this limit.
III
Telescoping series, the harmonic series and properties. Find the sum or show divergence.
- 1 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{(n+2)(n+3)}$
- 2 $\displaystyle\sum_{n=1}^{\infty}\frac{4}{4n^2-1}$
- 3 $\displaystyle\sum_{n=1}^{\infty}\left(\frac1{\sqrt n}-\frac1{\sqrt{n+1}}\right)$
- 4 $\displaystyle\sum_{n=1}^{\infty}\ln\frac{n+1}{n}$
- 5 $\displaystyle\sum_{n=1}^{\infty}\left(\cos\frac1n-\cos\frac1{n+1}\right)$
- 6 $\displaystyle\sum_{n=1}^{\infty}\left(\frac{2}{n(n+1)}-3\left(\frac14\right)^n\right)$
- 7 $\displaystyle\sum_{n=1}^{\infty}\left(\frac1n+\frac1{2^n}\right)$
- 8 $\displaystyle\sum_{n=1}^{\infty}\frac1{5n}$
- 9 The partial sums of a series are $S_N=\dfrac{3N}{N+1}$. Find $a_1$, a formula for $a_n$, and $\sum_{n=1}^{\infty}a_n$.
Numerical check: partial sums $S_N$ for large $N$
๐ก Show solutionsโถ
1 $\dfrac1{(n+2)(n+3)}=\dfrac1{n+2}-\dfrac1{n+3}$, so $S_N=\dfrac13-\dfrac1{N+3}\to\dfrac13$.
2 $\dfrac4{(2n-1)(2n+1)}=\dfrac2{2n-1}-\dfrac2{2n+1}$, so $S_N=2-\dfrac2{2N+1}\to2$.
3 $S_N=1-\dfrac1{\sqrt{N+1}}\to1$.
4 $\ln\frac{n+1}n=\ln(n+1)-\ln n$, so $S_N=\ln(N+1)\to\infty$: diverges.
5 $S_N=\cos1-\cos\dfrac1{N+1}\to\cos1-\cos0=\cos1-1\approx-0.4597$ (continuity of cosine).
6 $\displaystyle\sum\frac2{n(n+1)}=2$ and $\displaystyle\sum_{n=1}^\infty3\left(\tfrac14\right)^n=3\cdot\frac{1/4}{3/4}=1$; both converge, so the sum is $2-1=1$.
7 Divergent ($\sum\frac1n$) plus convergent ($\sum\frac1{2^n}$): diverges.
8 $\tfrac15\sum\tfrac1n$ โ a nonzero multiple of the harmonic series: diverges.
9 $a_1=S_1=\tfrac32$. For $n\ge2$, $a_n=S_n-S_{n-1}=\dfrac{3n}{n+1}-\dfrac{3(n-1)}{n}=\dfrac{3n^2-3(n^2-1)}{n(n+1)}=\dfrac{3}{n(n+1)}$, which also gives $\tfrac32$ at $n=1$. Sum $=\lim S_N=3$.
IV
Applications and modelling. Give units in every answer.
- 1 Sport. A basketball is dropped from $2$ m and always rebounds to $60\%$ of the height it fell from. Find the total vertical distance travelled.
- 2 For the ball in part 1, a fall (or rise) through height $h$ takes $\sqrt{2h/g}$ seconds, with $g=9.8$ m/sยฒ. Show the total time is $\sqrt{\tfrac{4}{9.8}}\left(1+\dfrac{2\sqrt{0.6}}{1-\sqrt{0.6}}\right)$ and evaluate it. Does the ball stop bouncing in finite time?
- 3 Economics โ the multiplier effect. A government injects 50 million euros into the economy. Every recipient spends $75\%$ of what they receive (and saves the rest), and that spending is received by others, who again spend $75\%$, and so on. Find the total spending generated, including the original 50 million.
- 4 Medicine. A patient takes $250$ mg of an antibiotic every $12$ hours; $30\%$ of the drug present remains after $12$ hours. Find the long-run amount just after a dose and just before a dose.
- 5 Fractals. The Koch snowflake starts with an equilateral triangle of area $1$. At step $n$ ($n\ge1$), $3\cdot4^{n-1}$ new small triangles are added, each of area $\left(\tfrac19\right)^n$. Find the total area of the snowflake. The perimeter after step $n$ is $3\left(\tfrac43\right)^n$ (side of the first triangle $1$) โ what happens to it?
- 6 Zeno revisited. Achilles runs at $10$ m/s and a tortoise at $1$ m/s with a $100$ m head start. Achilles needs $10$ s to reach the tortoise's starting point, by which time it has moved $10$ m; then $1$ s to cover that, and so on. Add up these times, and check your answer by elementary algebra.
๐ก Show solutionsโถ
1 $D=2+2\left(1.2+1.2(0.6)+1.2(0.6)^2+\cdots\right)=2+2\cdot\dfrac{1.2}{0.4}=2+6=8$ m.
2 The first fall takes $t_0=\sqrt{4/9.8}\approx0.6389$ s. The $k$th rebound height is $2(0.6)^k$, so each rise and each fall takes $t_0(\sqrt{0.6})^k$. Total $T=t_0\left(1+2\sum_{k=1}^\infty(\sqrt{0.6})^k\right)=t_0\left(1+\dfrac{2\sqrt{0.6}}{1-\sqrt{0.6}}\right)\approx0.6389\times7.873\approx5.03$ s. Yes: infinitely many bounces happen in about $5$ seconds.
3 $50+50(0.75)+50(0.75)^2+\cdots=\dfrac{50}{1-0.75}=200$ million euros. The multiplier is $\dfrac1{1-0.75}=4$.
4 Just after a dose: $250\left(1+0.3+0.3^2+\cdots\right)=\dfrac{250}{0.7}\approx357.1$ mg. Just before: $0.3\times357.1\approx107.1$ mg.
5 Added area $\displaystyle\sum_{n=1}^\infty3\cdot4^{n-1}\left(\tfrac19\right)^n=\sum_{n=1}^\infty\frac13\left(\frac49\right)^{n-1}=\frac{1/3}{1-4/9}=\frac35$. Total area $1+\tfrac35=\tfrac85$. The perimeter $3\left(\tfrac43\right)^n\to\infty$: a finite area enclosed by an infinitely long curve.
6 $T=10+1+0.1+\cdots=\dfrac{10}{1-0.1}=\dfrac{100}9\approx11.11$ s. Algebra: Achilles gains $9$ m/s on a $100$ m gap, so he catches up after $\tfrac{100}{9}$ s, at $\tfrac{1000}{9}\approx111.1$ m from his start. โ Zeno's infinitely many stages fit into a finite time.
๐ฌ ResearchโTeaching NexusResearch-tutored ยท optional, about 30 minutes
Research question. How should the dose and the dosing interval of a medicine be chosen so that the amount in the body settles inside its therapeutic window โ high enough to work, low enough to be safe?
Why it matters. Hospital pharmacists around the world design antibiotic, anti-epileptic and pain-relief regimens this way. Too low a trough lets bacteria recover and encourages resistance; too high a peak causes side effects. The geometric series is the mathematical core of these "steady-state" calculations.
What to do.
- Sample case: a drug has a half-life of $6$ hours, so after $\tau$ hours a fraction $f=2^{-\tau/6}$ remains. Its window is $300$โ$900$ mg.
- For $500$ mg every $\tau=8$ hours, compute $f$, then the amount just after the $n$th dose, $P_n=500\,\dfrac{1-f^n}{1-f}$, for $n=1,\dots,6$.
- Find the steady-state peak $P_\infty=\dfrac{500}{1-f}$ and trough $fP_\infty$. Are both inside the window?
- Now try $750$ mg every $12$ hours (the same total per day). Compute the new peak and trough and compare.
- How many doses until the peak is within $5\%$ of its steady value? Solve $f^n\lt0.05$.
Tools: Calculator or spreadsheet; optionally plot the amount against time (exponential decay between doses).
Hand in: A half-page recommendation to a ward pharmacist comparing the two regimens, with your numbers.
Going further. Look up a "loading dose": what first dose puts the patient at the steady-state peak immediately? (Answer in terms of $P_\infty$.)
Instructor note: $f=2^{-8/6}\approx0.397$; peak $\approx829$ mg, trough $\approx329$ mg โ inside the window. With $750$ mg every $12$ h, $f=0.25$, peak $1000$ mg and trough $250$ mg โ both outside. Within $5\%$ after $4$ doses for the 8-hour regimen ($0.397^4\approx0.025$, while $0.397^3\approx0.063$). Loading dose $=P_\infty$.
๐งช Self-check quiz
0/5 correct
Q1. $\displaystyle\sum_{n=0}^{\infty}\left(\frac23\right)^n=$
Q2. If $\displaystyle\lim_{n\to\infty}a_n=0$, then $\sum a_n$โฆ
Q3. $\displaystyle\sum_{n=1}^{\infty}\frac1{n(n+1)}=$
Q4. As a fraction, $0.\overline{27}=0.272727\ldots=$
Q5. A ball dropped from $6$ m always rebounds to half the height it fell. The total vertical distance it travels is