Endless lists of numbers $a_1,a_2,a_3,\dots$ โ what it means for them to settle on a limit $L$, and how to prove that they do.
A doctor prescribes a daily tablet; each day the body clears part of the drug, and a new dose is added. A bank compounds interest $n$ times a year, and we ask what happens as $n$ grows. A calculator finds $\sqrt5$ by repeating the step $x\mapsto\tfrac12\left(x+\tfrac5x\right)$. In each case the result is not one number but an endless list of numbers โ a sequence โ and the real question is: where is the list heading?
Sequences are the foundation for the rest of this chapter: an infinite series is defined as the limit of the sequence of its partial sums. In this section we learn to describe sequences, define their limits precisely, and develop the tools โ limit laws, L'Hรดpital's rule, the squeeze theorem and the Monotone Convergence Theorem โ that decide whether a sequence converges.
Defining a sequence
๐ Definition 8.1 โ Sequence
A sequence is an ordered, infinite list of numbers $$a_1,\ a_2,\ a_3,\ \dots,\ a_n,\ \dots$$ written $\{a_n\}$ or $\{a_n\}_{n=1}^{\infty}$. Each $a_n$ is a term and $n$ is its index. Formally, a sequence is a function whose domain is the positive integers: $a_n=a(n)$. (Sometimes the index starts at $0$ or another integer; we will say so when it does.) A sequence can be given
explicitly, by a formula $a_n=f(n)$ โ for example $a_n=\dfrac{n}{n+1}$ gives $\tfrac12,\tfrac23,\tfrac34,\dots$;
recursively, by one or more starting values and a rule that builds each term from earlier ones โ for example $a_1=1$, $a_{n+1}=2a_n+1$ gives $1,3,7,15,\dots$
Two families appear everywhere. An arithmetic sequence adds a fixed difference $d$ each time, $a_n=a_1+(n-1)d$. A geometric sequence multiplies by a fixed ratio $r$ each time, $a_n=a_1r^{\,n-1}$. The most famous recursive sequence is the Fibonacci sequence $F_1=F_2=1$, $F_{n+2}=F_{n+1}+F_n$, giving $1,1,2,3,5,8,13,\dots$
โ๏ธ Example 1 โ Listing terms and finding a formula
(a) Write the first five terms of $a_n=(-1)^n\dfrac{2n}{n+1}$.
(b) Write the first five terms of $a_1=2$, $a_{n+1}=3a_n-2$, and check that the explicit formula $a_n=3^{\,n-1}+1$ produces the same terms.
(c) Find an explicit formula for the sequence $\dfrac34,\ \dfrac97,\ \dfrac{27}{10},\ \dfrac{81}{13},\ \dots$
๐ก Show solutionโถ
(a) Substitute $n=1,\dots,5$; the factor $(-1)^n$ makes odd-numbered terms negative: $$a_1=-\frac22=-1,\quad a_2=\frac43,\quad a_3=-\frac64=-\frac32,\quad a_4=\frac85,\quad a_5=-\frac{10}6=-\frac53.$$
(b) Apply the rule repeatedly: $a_1=2$, $a_2=3(2)-2=4$, $a_3=3(4)-2=10$, $a_4=3(10)-2=28$, $a_5=3(28)-2=82$. The formula gives $3^0+1=2$, $3^1+1=4$, $3^2+1=10$, $3^3+1=28$, $3^4+1=82$. โ (In general, if $a_n=3^{n-1}+1$ then $3a_n-2=3^n+3-2=3^n+1=a_{n+1}$, so the formula is right for every $n$.)
(c) Numerators $3,9,27,81$ are $3^n$. Denominators $4,7,10,13$ increase by $3$, an arithmetic sequence $4+3(n-1)=3n+1$. So $$a_n=\frac{3^n}{3n+1}.$$
Look at $a_n=\dfrac{2n+1}{n+3}$: its terms are $0.75$, $1$, $1.1667$, $1.2857$, โฆ, $a_{100}=1.9515$, $a_{1000}=1.9950$. They get closer and closer to $2$ and stay close. Informally, $a_n\to L$ means "$a_n$ is as close to $L$ as we like, provided $n$ is large enough". To make "as close as we like" precise we name the allowed error $\varepsilon$ (epsilon), and to make "large enough" precise we name a cut-off index $N$.
๐ Definition 8.2 โ Convergent and divergent sequences
A sequence $\{a_n\}$ converges to the real number $L$, written $$\lim_{n\to\infty}a_n=L\qquad\text{or}\qquad a_n\to L,$$ if for every $\varepsilon\gt0$ there is an integer $N$ such that $$|a_n-L|\lt\varepsilon\quad\text{for all } n\gt N.$$ Geometrically: whatever band $(L-\varepsilon,\ L+\varepsilon)$ you draw around $L$, from some point on all the terms lie inside it. A sequence that does not converge diverges. If for every $M\gt0$ there is an $N$ with $a_n\gt M$ for all $n\gt N$, we say $a_n$ diverges to infinity and write $a_n\to\infty$ (similarly $a_n\to-\infty$).
Notice the order of the quantifiers: the challenger picks $\varepsilon$ first, then we must find a suitable $N$. A smaller $\varepsilon$ usually forces a larger $N$. The limit, if it exists, is unique, and changing finitely many terms never affects it.
โ๏ธ Example 2 โ Using the ฮตโN definition
Prove that $\displaystyle\lim_{n\to\infty}\frac{2n+1}{n+3}=2$. Then find the smallest valid $N$ when $\varepsilon=0.1$ and when $\varepsilon=0.01$.
๐ก Show solutionโถ
Measure the error $$|a_n-2|=\left|\frac{2n+1-2(n+3)}{n+3}\right|=\left|\frac{-5}{n+3}\right|=\frac{5}{n+3}.$$
Solve for n $\dfrac5{n+3}\lt\varepsilon\iff n+3\gt\dfrac5\varepsilon\iff n\gt\dfrac5\varepsilon-3.$ So, given any $\varepsilon\gt0$, choose $N$ to be any integer with $N\ge\dfrac5\varepsilon-3$. Then every $n\gt N$ satisfies $|a_n-2|\lt\varepsilon$. By Definition 8.2, $a_n\to2$. โ
Numbers $\varepsilon=0.1$: $n\gt50-3=47$, so $N=47$. $\varepsilon=0.01$: $n\gt500-3=497$, so $N=497$.
Check $a_{497}=\dfrac{995}{500}=1.99$, error exactly $0.01$ โ not less than $\varepsilon$. $a_{498}=\dfrac{997}{501}\approx1.99002$, error $\approx0.00998\lt0.01$ โ. Interactive 1 shows this band.
Interactive 1 โ The ฮต-Band and the Cut-off N
The green band is $(L-\varepsilon,\ L+\varepsilon)$. Terms after the dashed line $n=N$ (blue) are all inside it; red dots are terms that escape the band. Shrink $\varepsilon$ (or press Play) and watch $N$ move right โ but for a convergent sequence some $N$ always exists. Use Zoom to magnify the band.
โ
$\varepsilon$โ
cut-off $N$โ
$a_{N+1}$โ
$|a_{N+1}-L|$โ
Proving limits from the definition is important for understanding, but in practice we use rules. Because a sequence is just a function on the integers, the limit laws for functions carry over.
๐ Theorem 8.1 โ Limit laws for sequences
Suppose $\displaystyle\lim_{n\to\infty}a_n=A$ and $\displaystyle\lim_{n\to\infty}b_n=B$, where $A$ and $B$ are real numbers, and let $c$ be a constant. Then
$$\lim(a_n\pm b_n)=A\pm B,\qquad \lim(c\,a_n)=cA,\qquad \lim(a_nb_n)=AB,$$
$$\lim\frac{a_n}{b_n}=\frac AB\ \text{ provided } B\ne0,\qquad \lim (a_n)^p=A^p\ \text{ provided } p\gt0 \text{ and } a_n\gt0.$$
๐ Theorem 8.2 โ Sequences defined by functions
Let $f$ be a function defined for all real $x\ge1$, and suppose $\displaystyle\lim_{x\to\infty}f(x)=L$ (where $L$ is a real number, $\infty$ or $-\infty$). If $a_n=f(n)$ for every integer $n\ge1$, then $\displaystyle\lim_{n\to\infty}a_n=L$.
Theorem 8.2 is what lets us use L'Hรดpital's rule on sequences: L'Hรดpital's rule needs a differentiable function of a real variable, so we apply it to $f(x)$ and then read off the limit of $a_n=f(n)$. (We never "differentiate with respect to $n$".)
If $a_n\to L$ and the function $g$ is continuous at $L$, then $g(a_n)\to g(L)$. In symbols, $\displaystyle\lim_{n\to\infty}g(a_n)=g\!\left(\lim_{n\to\infty}a_n\right)$.
โ ๏ธ The converse of Theorem 8.2 is false
If $a_n=f(n)$ converges, $f(x)$ need not have a limit as $x\to\infty$. For example $a_n=\sin(\pi n)=0$ for every integer $n$, so $a_n\to0$, but $f(x)=\sin(\pi x)$ oscillates forever. A sequence only "sees" the function at the integers.
โ๏ธ Example 3 โ Limits by algebra, L'Hรดpital's rule and continuity
Find each limit or show that the sequence diverges.
$$\text{(a) } a_n=\frac{3n^2-n}{5n^2+2}\qquad \text{(b) } a_n=\frac{\ln n}{n}\qquad \text{(c) } a_n=\frac{n^2}{e^n}\qquad \text{(d) } a_n=\sqrt{\frac{4n+1}{n+2}}$$
๐ก Show solutionโถ
(a) Divide numerator and denominator by the highest power, $n^2$: $$\lim_{n\to\infty}\frac{3-\frac1n}{5+\frac2{n^2}}=\frac{3-0}{5+0}=\frac35.$$
(b) Let $f(x)=\dfrac{\ln x}{x}$, an $\dfrac\infty\infty$ form. By L'Hรดpital's rule $\displaystyle\lim_{x\to\infty}\frac{\ln x}{x}=\lim_{x\to\infty}\frac{1/x}{1}=0$. By Theorem 8.2, $\dfrac{\ln n}{n}\to0$.
(c) $f(x)=\dfrac{x^2}{e^x}$ is $\dfrac\infty\infty$; apply L'Hรดpital twice: $$\lim_{x\to\infty}\frac{x^2}{e^x}=\lim_{x\to\infty}\frac{2x}{e^x}=\lim_{x\to\infty}\frac{2}{e^x}=0.$$ So $a_n\to0$: exponentials beat powers.
(d) Inside: $\dfrac{4n+1}{n+2}=\dfrac{4+1/n}{1+2/n}\to4$. Since $g(x)=\sqrt x$ is continuous at $4$, Theorem 8.3 gives $a_n\to\sqrt4=2$.
Some sequences, such as $\dfrac{\cos n}{n}$, wobble too irregularly for L'Hรดpital's rule. If we can trap such a sequence between two simpler sequences that share a limit, it has no choice but to follow them.
๐ Theorem 8.4 โ Squeeze theorem for sequences
Suppose $a_n\le b_n\le c_n$ for all $n\ge n_0$ (some fixed integer $n_0$), and $\displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty}c_n=L$. Then $\displaystyle\lim_{n\to\infty}b_n=L$. Corollary. If $|a_n|\to0$ then $a_n\to0$ (squeeze $a_n$ between $-|a_n|$ and $|a_n|$).
๐ Theorem 8.5 โ Six standard limits
Geometric sequences$\displaystyle\lim_{n\to\infty}r^n=\begin{cases}0,&|r|\lt1\\1,&r=1\end{cases}$; $\{r^n\}$ diverges if $r\le-1$ or $r\gt1$ (to $\infty$ when $r\gt1$).
nth root of n$\displaystyle\lim_{n\to\infty}n^{1/n}=1$
nth root of a constant$\displaystyle\lim_{n\to\infty}x^{1/n}=1$ for every fixed $x\gt0$
The exponential limit$\displaystyle\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x$ for every real $x$
Factorials beat powers$\displaystyle\lim_{n\to\infty}\frac{x^n}{n!}=0$ for every real $x$
$n^{1/n}$ and $x^{1/n}$: $\ln\left(n^{1/n}\right)=\dfrac{\ln n}{n}\to0$ (Example 3(b)), so by continuity of $e^x$, $n^{1/n}=e^{\ln n/n}\to e^0=1$. Likewise $\ln\left(x^{1/n}\right)=\dfrac{\ln x}{n}\to0$.
$(1+x/n)^n$: for $x\ne0$, $\ln\left(1+\tfrac xn\right)^n=\dfrac{\ln(1+x/n)}{1/n}$, a $\tfrac00$ form. With $t=\tfrac1n$ and L'Hรดpital, $\displaystyle\lim_{t\to0^+}\frac{\ln(1+xt)}{t}=\lim_{t\to0^+}\frac{x}{1+xt}=x$, so the sequence tends to $e^x$.
$n!/n^n$: $\dfrac{n!}{n^n}=\dfrac1n\cdot\dfrac2n\cdots\dfrac nn\le\dfrac1n$, since every factor after the first is at most $1$. So $0\lt\dfrac{n!}{n^n}\le\dfrac1n$ and the squeeze theorem gives $0$.
$x^n/n!$: proved in Example 5 for $x=3$; the same argument works for any $x$.
โ๏ธ Example 4 โ Squeezing and standard limits
Find each limit or show that the sequence diverges.
$$\text{(a) } \frac{\cos n}{n}\qquad\text{(b) } \left(1-\frac2n\right)^n\qquad\text{(c) } \left(2^n+3^n\right)^{1/n}\qquad\text{(d) } (-1)^n$$
๐ก Show solutionโถ
(a) Since $-1\le\cos n\le1$, $\ -\dfrac1n\le\dfrac{\cos n}{n}\le\dfrac1n$. Both outer sequences tend to $0$, so $\dfrac{\cos n}{n}\to0$ by the squeeze theorem.
(b) The exponential limit with $x=-2$: $\left(1+\dfrac{-2}{n}\right)^n\to e^{-2}\approx0.1353$.
(c) The larger base dominates. Since $3^n\le2^n+3^n\le2\cdot3^n$, taking $n$th roots (an increasing operation on positive numbers) gives $$3\le\left(2^n+3^n\right)^{1/n}\le2^{1/n}\cdot3.$$ By the standard limit $2^{1/n}\to1$, the right side tends to $3$; by the squeeze theorem the limit is $3$.
(d) The terms are $-1,1,-1,1,\dots$ Suppose $(-1)^n\to L$. Take $\varepsilon=1$: for large $n$ both $|1-L|\lt1$ and $|-1-L|\lt1$, so $2=|1-(-1)|\le|1-L|+|L-(-1)|\lt2$, a contradiction. The sequence diverges (it oscillates). Interactive 1, last option, shows why no $N$ works.
Monotone and bounded sequences
Often we can show that a sequence converges without knowing its limit in advance โ essential for recursive sequences, whose terms we cannot write in closed form.
๐ Definition 8.3 โ Monotone and bounded sequences
A sequence $\{a_n\}$ is increasing if $a_n\le a_{n+1}$ for all $n$, and decreasing if $a_n\ge a_{n+1}$ for all $n$ (strictly, if the inequalities are strict). It is monotone if it is increasing or decreasing. It is bounded above if there is a number $M$ with $a_n\le M$ for all $n$, bounded below if there is an $m$ with $a_n\ge m$ for all $n$, and bounded if it is both.
๐ Theorem 8.6 โ Convergent sequences are bounded
If $\{a_n\}$ converges, then $\{a_n\}$ is bounded. Equivalently: an unbounded sequence diverges. (The converse is false: $(-1)^n$ is bounded but diverges.)
๐ Theorem 8.7 โ Monotone Convergence Theorem
If $\{a_n\}$ is bounded and monotone, then $\{a_n\}$ converges. More precisely: an increasing sequence that is bounded above converges (to the least number $L$ with $a_n\le L$ for all $n$), and a decreasing sequence that is bounded below converges. The conclusion still holds if the sequence is only monotone from some index $n_0$ onwards.
Why should this be true? An increasing sequence can only move right on the number line; if a wall $M$ blocks it, the terms must pile up somewhere at or before the wall. This depends on the completeness of the real numbers โ the real line has no gaps. (In the rational numbers alone the theorem fails: the increasing, bounded sequence $1,\ 1.4,\ 1.41,\ 1.414,\dots$ is heading for $\sqrt2$, which is not rational.)
๐งญ Three ways to show a sequence is monotone
Difference: show $a_{n+1}-a_n\ge0$ (increasing) or $\le0$ (decreasing).
Ratio (for positive terms, especially with factorials and powers): show $\dfrac{a_{n+1}}{a_n}\ge1$ or $\le1$.
Derivative: if $a_n=f(n)$, show $f^\prime(x)\ge0$ or $\le0$ for $x\ge1$.
For recursive sequences use induction: prove the bound and the monotonicity step by step, then find the limit $L$ by letting $n\to\infty$ on both sides of the recursion โ which is allowed only after you know the limit exists.
โ๏ธ Example 5 โ Monotone and bounded: $3^n/n!$
Show that $a_n=\dfrac{3^n}{n!}$ is eventually decreasing and bounded below, and find its limit.
๐ก Show solutionโถ
Ratio $$\frac{a_{n+1}}{a_n}=\frac{3^{n+1}}{(n+1)!}\cdot\frac{n!}{3^n}=\frac{3}{n+1}.$$ This is $\le1$ exactly when $n\ge2$. So $a_2\ge a_3\ge a_4\ge\cdots$: the sequence is decreasing from $n=2$ on. (The first terms are $3,\ 4.5,\ 4.5,\ 3.375,\ 2.025,\ 1.0125,\dots$ โ it rises before it falls.)
Bounded Every term is positive, so $a_n\gt0$: bounded below by $0$. By the Monotone Convergence Theorem (monotone from $n_0=2$), $a_n\to L$ for some $L\ge0$.
Find L From the ratio, $a_{n+1}=\dfrac{3}{n+1}\,a_n$. Let $n\to\infty$: the left side tends to $L$ and the right side to $0\cdot L=0$. So $L=0$: $\dfrac{3^n}{n!}\to0$.
Let $a_1=1$ and $a_{n+1}=\sqrt{2+a_n}$. Show that $\{a_n\}$ converges and find its limit.
๐ก Show solutionโถ
Explore $a_1=1$, $a_2=\sqrt3\approx1.7321$, $a_3\approx1.9319$, $a_4\approx1.9829$, $a_5\approx1.9957$. It looks increasing, with limit $2$.
Bounded above by 2 By induction. $a_1=1\lt2$. If $a_k\lt2$, then $a_{k+1}=\sqrt{2+a_k}\lt\sqrt{2+2}=2$. Hence $a_n\lt2$ for all $n$. Also every $a_n\ge1\gt0$.
Increasing For positive terms, $a_{n+1}\gt a_n\iff2+a_n\gt a_n^2\iff(2-a_n)(1+a_n)\gt0$, which is true because $0\lt a_n\lt2$. So the sequence is increasing.
Limit Bounded and monotone, so $a_n\to L$ by Theorem 8.7. Since $a_{n+1}\to L$ as well and $\sqrt{\phantom{x}}$ is continuous, $L=\sqrt{2+L}$, so $L^2-L-2=0$, $(L-2)(L+1)=0$. As $L\ge1$, $\boxed{L=2}$.
Picture Interactive 2 draws this as a cobweb: the limit is where $y=\sqrt{2+x}$ meets $y=x$.
โ ๏ธ Find the limit only after proving it exists
Setting $L=f(L)$ assumes a limit exists. For $a_1=1$, $a_{n+1}=2a_n+1$ the same trick gives $L=2L+1$, so "$L=-1$" โ nonsense, since the terms $1,3,7,15,\dots$ grow without bound. First prove convergence (e.g. with the Monotone Convergence Theorem); then solve $L=f(L)$ and pick the root consistent with your bounds.
Interactive 2 โ Cobweb Diagrams for Recursive Sequences
For $a_{n+1}=f(a_n)$: go up from $a_n$ to the curve $y=f(x)$ (that height is $a_{n+1}$), then across to the line $y=x$ to move $a_{n+1}$ back onto the $x$-axis. Repeat. A staircase or spiral closing in on a crossing point means convergence to a solution of $L=f(L)$; drag the starting value $a_1$ to explore.
โ
$n$โ
$a_n$โ
$|a_n-a_{n-1}|$โ
long-run behaviourโ
Exercises
I
Terms and formulas. In 1โ4 write the requested terms; in 5โ8 find an explicit formula $a_n$ (starting at $n=1$).
1 The first four terms of $a_n=\dfrac{2n-1}{n+1}$.
2 The first four terms of $a_n=\dfrac{(-1)^n}{2^n}$.
3 The first four terms of $a_1=3$, $a_{n+1}=2a_n-1$; guess an explicit formula.
4 The first five terms of $a_1=a_2=1$, $a_{n+2}=a_{n+1}+a_n$.
8 Each term is one less than a perfect square $1,4,9,16,25$: $a_n=n^2-1$.
II
Determine whether each sequence converges or diverges. If it converges, find its limit; name the tool you use (algebra, L'Hรดpital, continuity, squeeze, standard limit).
1 $a_n=\dfrac{5n+2}{3n-1}$
2 $a_n=\dfrac{n^2+1}{2n^3-n}$
3 $a_n=\dfrac{n^3}{n^2+4}$
4 $a_n=\sqrt{n^2+n}-n$
5 $a_n=(-1)^n\dfrac{n}{n+1}$
6 $a_n=\dfrac{\sin n}{\sqrt n}$
7 $a_n=\dfrac{\ln n}{\sqrt n}$
8 $a_n=n^3e^{-n}$
9 $a_n=\left(1+\dfrac4n\right)^n$
10 $a_n=(3n)^{1/n}$
11 $a_n=\tan^{-1}n$
12 $a_n=n\sin\dfrac1n$
13 $a_n=\dfrac{10^n}{n!}$
14 $a_n=\left(-\dfrac34\right)^n$
15 $a_n=\dfrac{3^{n+1}}{4^n}$
16 $a_n=(1.01)^n$
Numerical check: evaluate $a_n$ far down the sequence
๐ก Show solutionsโถ
1 Divide by $n$: $\dfrac{5+2/n}{3-1/n}\to\dfrac53$. Converges to $\tfrac53$.
2 Divide by $n^3$: $\dfrac{1/n+1/n^3}{2-1/n^2}\to\dfrac02=0$. Converges to $0$.
3 $\dfrac{n^3}{n^2+4}=\dfrac{n}{1+4/n^2}\to\infty$. Diverges to $\infty$ (unbounded, Theorem 8.6).
4 Multiply by the conjugate: $\sqrt{n^2+n}-n=\dfrac{(n^2+n)-n^2}{\sqrt{n^2+n}+n}=\dfrac{n}{\sqrt{n^2+n}+n}=\dfrac{1}{\sqrt{1+1/n}+1}\to\dfrac12$.
5 $\dfrac n{n+1}\to1$, so even-numbered terms approach $1$ and odd-numbered terms approach $-1$. Two different targets: diverges.
6 Squeeze: $-\dfrac1{\sqrt n}\le\dfrac{\sin n}{\sqrt n}\le\dfrac1{\sqrt n}$ and $\dfrac1{\sqrt n}\to0$. Converges to $0$.
7 L'Hรดpital on $f(x)=\dfrac{\ln x}{\sqrt x}$: $\displaystyle\lim_{x\to\infty}\frac{1/x}{1/(2\sqrt x)}=\lim_{x\to\infty}\frac{2}{\sqrt x}=0$. Converges to $0$ (slowly: $a_{10^6}\approx0.0138$).
8 $f(x)=\dfrac{x^3}{e^x}$; L'Hรดpital three times gives $\displaystyle\lim\frac{6}{e^x}=0$. Converges to $0$.
9 Standard limit with $x=4$: $\left(1+\tfrac4n\right)^n\to e^4\approx54.598$.
10 $(3n)^{1/n}=3^{1/n}\cdot n^{1/n}\to1\cdot1=1$ (standard limits and the product law).
11 $\displaystyle\lim_{x\to\infty}\tan^{-1}x=\frac\pi2$, so by Theorem 8.2, $a_n\to\dfrac\pi2\approx1.5708$.
12 Write $n\sin\dfrac1n=\dfrac{\sin(1/n)}{1/n}$. With $t=\tfrac1n\to0^+$: $\dfrac{\sin t}{t}\to1$. Converges to $1$.
13 Standard limit $\dfrac{x^n}{n!}\to0$ with $x=10$. (The terms first grow โ $a_{10}\approx2756$ โ and only then collapse.) Converges to $0$.
14 Geometric with $r=-\tfrac34$, $|r|\lt1$: converges to $0$.
15 $\dfrac{3^{n+1}}{4^n}=3\left(\tfrac34\right)^n\to3\cdot0=0$. Converges to $0$.
16 Geometric with $r=1.01\gt1$: $(1.01)^n\to\infty$. Diverges โ though slowly: $(1.01)^{100}\approx2.70$.
III
Monotone, bounded and recursive sequences. In 1โ5 decide whether the sequence is monotone and whether it is bounded, then state whether it converges. In 6โ8 prove convergence and find the limit.
1 $a_n=\dfrac{n}{2n+1}$
2 $a_n=\dfrac{1}{n+\ln n}$
3 $a_n=ne^{-n}$
4 $a_n=(-1)^n$. Does this contradict the Monotone Convergence Theorem?
5 $a_n=\dfrac{2^n}{n!}$
6 $a_1=0$, $a_{n+1}=\sqrt{6+a_n}$
7 $a_1=1$, $a_{n+1}=3-\dfrac1{a_n}$
8 $a_1=1$, $a_{n+1}=\tfrac12a_n+1$. Also find an explicit formula.
2 $n+\ln n$ is increasing and positive, so its reciprocal is decreasing; $0\lt a_n\le a_1=1$. Bounded and monotone: converges (to $0$).
3 $f(x)=xe^{-x}$ has $f^\prime(x)=(1-x)e^{-x}\le0$ for $x\ge1$: decreasing. $0\lt a_n\le a_1=e^{-1}\approx0.368$. Converges (to $0$, by L'Hรดpital on $x/e^x$).
4 Bounded ($|a_n|=1$) but not monotone, and divergent (Example 4(d)). No contradiction: the theorem needs both hypotheses, and monotonicity fails here.
5 $\dfrac{a_{n+1}}{a_n}=\dfrac{2}{n+1}\le1$ for all $n\ge1$ (equality at $n=1$: $a_1=a_2=2$), so decreasing; $0\lt a_n\le2$. Converges; as in Example 5, $L=0\cdot L=0$.
6 Induction: $a_n\lt3$ (since $a_1=0\lt3$ and $a_k\lt3\Rightarrow a_{k+1}\lt\sqrt9=3$), and $a_n\ge0$. Increasing: $a_{n+1}\gt a_n\iff6+a_n\gt a_n^2\iff(3-a_n)(2+a_n)\gt0$, true for $0\le a_n\lt3$. So $a_n\to L$ with $L=\sqrt{6+L}$, $L^2-L-6=(L-3)(L+2)=0$, $L=3$ (as $L\ge0$).
7 Terms $1,\ 2,\ 2.5,\ 2.6,\ 2.6154,\dots$ Let $g(x)=3-\tfrac1x$, which is increasing for $x\gt0$, and let $L^*=\tfrac{3+\sqrt5}2$, the positive root of $x=3-\tfrac1x$. Induction: $1\le a_n\lt L^*$ (because $a_n\lt L^*\Rightarrow a_{n+1}=g(a_n)\lt g(L^*)=L^*$), and $a_n\lt a_{n+1}$ (true for $n=1$; if $a_n\lt a_{n+1}$ then $g(a_n)\lt g(a_{n+1})$, i.e. $a_{n+1}\lt a_{n+2}$). Increasing and bounded, so $L=3-\tfrac1L$, $L^2-3L+1=0$, $L=\dfrac{3+\sqrt5}{2}\approx2.618$ (the other root $\approx0.382$ is below $a_1$).
8 Terms $1,\ 1.5,\ 1.75,\ 1.875,\dots$ Induction gives $a_n\lt2$ and $a_{n+1}-a_n=1-\tfrac12a_n\gt0$, so it converges, and $L=\tfrac12L+1$ gives $L=2$. Explicitly, $2-a_{n+1}=\tfrac12(2-a_n)$, so $2-a_n=\left(\tfrac12\right)^{n-1}(2-a_1)$ and $a_n=2-2^{1-n}\to2$ โ.
IV
Applications, proofs and concepts.
1Applied (medicine). A patient takes a 50 mg tablet every morning. Over 24 hours the body eliminates 40% of the drug present. Let $Q_n$ be the amount (mg) just after the $n$th tablet, so $Q_1=50$ and $Q_{n+1}=0.6\,Q_n+50$. Find $Q_2,Q_3,Q_4$.
2 For the sequence in part 1, verify that $Q_n=125-75(0.6)^{n-1}$ and find the long-run amount $\lim Q_n$.
3Applied (finance). 1000 euros is invested for one year at 6% annual interest compounded $n$ times a year, giving $A_n=1000\left(1+\tfrac{0.06}{n}\right)^n$ euros. Compute $A_1,A_4,A_{12},A_{365}$ and $\lim A_n$ (continuous compounding).
4 Use the ฮตโN definition to prove $\displaystyle\lim_{n\to\infty}\frac{3n-1}{n+2}=3$, and find the smallest $N$ that works for $\varepsilon=0.001$.
5Applied (computing). Newton's method for $\sqrt5$ uses $x_1=2$, $x_{n+1}=\tfrac12\left(x_n+\tfrac5{x_n}\right)$. Compute $x_2,x_3,x_4$. Assuming the sequence converges to a positive limit, find it.
6Concepts. True or false? Give a reason or a counterexample. (a) Every bounded sequence converges. (b) If $a_n\to L$ then $a_{n+1}\to L$. (c) If $|a_n|\to0$ then $a_n\to0$. (d) If $|a_n|\to1$ then $a_n\to1$. (e) If $a_n\to0$ and $\{b_n\}$ is bounded, then $a_nb_n\to0$.
2 $n=1$: $125-75=50$ โ. If $Q_n=125-75(0.6)^{n-1}$ then $0.6Q_n+50=75-45(0.6)^{n-1}+50=125-75(0.6)^n=Q_{n+1}$ โ. Since $(0.6)^{n-1}\to0$, $Q_n\to125$ mg. (Check with the fixed point: $L=0.6L+50\Rightarrow L=125$.)
3 $A_1=1060.00$, $A_4\approx1061.36$, $A_{12}\approx1061.68$, $A_{365}\approx1061.83$ euros. By the exponential limit with $x=0.06$: $\lim A_n=1000e^{0.06}\approx1061.84$ euros. More frequent compounding helps, but only up to a ceiling.
4 $\left|\dfrac{3n-1}{n+2}-3\right|=\dfrac{7}{n+2}\lt\varepsilon\iff n\gt\dfrac7\varepsilon-2$; take $N\ge\dfrac7\varepsilon-2$. For $\varepsilon=0.001$: $n\gt6998$, so $N=6998$ (at $n=6998$ the error is exactly $0.001$).
5 $x_2=\tfrac12(2+2.5)=2.25$, $x_3=\tfrac12\left(2.25+\tfrac5{2.25}\right)\approx2.236111$, $x_4\approx2.236068$. A positive limit satisfies $L=\tfrac12\left(L+\tfrac5L\right)$, i.e. $L^2=5$, so $L=\sqrt5\approx2.2360680$. Already $x_4$ is correct to 7 significant figures.
6 (a) False $(-1)^n$. (b) True it is the same list with the first term removed. (c) True corollary of the squeeze theorem. (d) False $(-1)^n$ has $|a_n|=1$. (e) True if $|b_n|\le M$ then $0\le|a_nb_n|\le M|a_n|\to0$; squeeze.
๐ฌ ResearchโTeaching NexusResearch-tutored ยท optional, about 30 minutes
Research question. When does a fish population governed by the recursive model $x_{n+1}=r\,x_n(1-x_n)$ settle to a steady level, and when does it oscillate or behave chaotically?
Why it matters. Fisheries scientists โ from the North Atlantic cod stocks to tuna and kingfish fisheries in the Indian Ocean โ use recursive population models to set catch limits. If the model predicts oscillations, a catch quota that is safe one year may collapse the stock the next. The logistic map ($x_n$ = population as a fraction of the habitat's capacity, $r$ = growth factor) is the simplest model showing all three behaviours.
What to do.
Use Interactive 2 (logistic option) or a spreadsheet: put $x_1=0.2$ in one cell and $=r\cdot x_n(1-x_n)$ below it, copied down 100 rows.
For $r=2.5$ and $r=2.8$, show the sequence converges, and check that the limit equals the fixed point $1-\tfrac1r$.
For $r=3.2$ show that $x_n$ does not converge: the odd- and even-numbered terms approach two different values. Find them.
Increase $r$ in steps of $0.05$ from $3.0$ to $3.6$ and record the long-run pattern (1 value, 2 values, 4 values, โฆ). Then try $r=3.9$.
Explain, using the slope $f^\prime(L)$ of $f(x)=rx(1-x)$ at the fixed point, why convergence fails once $r\gt3$. (Hint: compute $|f^\prime(1-1/r)|=|2-r|$.)
Tools: Spreadsheet or Python; Interactive 2 on this page.
Hand in: A one-page note with a table of $r$ against long-run behaviour, one cobweb sketch, and a sentence of advice for a fisheries manager.
Going further. Look up "period doubling" and the Feigenbaum constant $\delta\approx4.669$, and compare with the $r$-values where your pattern doubled.
Instructor note: For $1\lt r\lt3$ the sequence converges to $1-\tfrac1r$ ($0.6$ for $r=2.5$; $\approx0.6429$ for $r=2.8$). At $r=3.2$ it alternates between $\approx0.5130$ and $\approx0.7995$. A 4-cycle appears near $r\approx3.449$ ($=1+\sqrt6$), and chaos sets in around $r\approx3.57$. The fixed point attracts exactly when $|2-r|\lt1$.