--- title: Sequences module: Infinite Sequences and Series moduleNumber: 9 lessonNumber: 1 order: 901 summary: > A sequence is a function on the positive integers, and its limit is defined almost exactly like a limit at infinity. The Limit Laws and Squeeze Theorem carry over from functions, monotonic and bounded sequences give a convergence criterion, and the Monotonic Sequence Theorem guarantees a limit exists without naming it. topics: [Infinite Sequences and Series] sources: - book: Stewart ref: "Ch. 11 โ€” Infinite Sequences and Series; ยง11.1 Sequences" --- A **sequence** is a list of numbers written in a definite order, $$ a_1, a_2, a_3, \ldots, a_n, \ldots $$ The number $a_1$ is the first term, $a_2$ the second, and $a_n$ the $n$th or **general term**. Every term has a successor, so the list never ends. Because each positive integer $n$ determines exactly one term $a_n$, a sequence is a _function whose domain is the set of positive integers_. The only reason we write $a_n$ instead of $f(n)$ is convention; nothing else about the function idea changes. The sequence $\{a_1, a_2, a_3, \ldots\}$ is denoted $\{a_n\}$ or $\{a_n\}_{n=1}^{\infty}$. A sequence can be given three ways: by a formula for the general term, by listing terms, or by a rule that has no simple closed form. - **By formula.** $a_n = \dfrac{n}{n+1}$ generates $\tfrac{1}{2}, \tfrac{2}{3}, \tfrac{3}{4}, \tfrac{4}{5}, \ldots$ - **By a recurrence.** The Fibonacci sequence $\{f_n\}$ is defined by $f_1 = 1$, $f_2 = 1$, and $f_n = f_{n-1} + f_{n-2}$ for $n \ge 3$, giving $1, 1, 2, 3, 5, 8, 13, 21, \ldots$ โ€” each term is the sum of the two before it. - **By description alone.** Let $a_n$ be the $n$th decimal digit of $e$; then $\{a_n\} = \{7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, \ldots\}$ is a perfectly well-defined sequence with no generating formula. The index need not start at $1$. The sequence $\{\sqrt{n-3}\}_{n=3}^{\infty}$ begins at $n = 3$, and $\{\cos(n\pi/6)\}_{n=0}^{\infty}$ begins at $n = 0$. ## The limit of a sequence Because a sequence is a function on the integers, it can be pictured either by plotting its terms on a number line or by plotting the points $(n, a_n)$ in the plane. The graph is a set of _isolated_ points, one above each integer. For $a_n = n/(n+1)$ the points climb toward the line $y = 1$: the gap $$ 1 - \frac{n}{n+1} = \frac{1}{n+1} $$ can be made as small as we like by taking $n$ large enough. We write $\lim_{n\to\infty} n/(n+1) = 1$. $$ % caption: The terms of $a_n = n/(n+1)$ are isolated points that climb toward % the line $y = L = 1$; past some index $N$ every term lies inside the band of % half-width $\varepsilon$ about $L$. \begin{tikzpicture}[scale=1.0, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black, ->] (0,0) -- (8.4,0) node[anchor=north east] {$n$}; \draw[black, ->] (0,0) -- (0,3.9) node[anchor=north east] {$a_n$}; % band about L=3 (scaled: value*3) \fill[acc!10] (0,2.55) rectangle (8,3.45); \draw[acc, dashed] (0,3) -- (8,3) node[anchor=south east, black] {$L = 1$}; \draw[black, dashed] (0,2.55) -- (8,2.55); \draw[black, dashed] (0,3.45) -- (7,3.45); \draw[black, <->] (7.4,2.55) -- (7.4,3.45) node[midway, anchor=west, black, font=\scriptsize] {band}; \foreach \n/\y in {1/1.5, 2/2.0, 3/2.25, 4/2.4, 5/2.5, 6/2.571, 7/2.625, 8/2.667} \fill (\n,\y) circle (2.4pt); \draw[black, dotted] (6,0) -- (6,2.571); \node[anchor=north, black!70] at (6,0) {$N$}; \end{tikzpicture} $$ The picture suggests the intuitive definition: $a_n \to L$ if the terms get and stay arbitrarily close to $L$. The precise version copies the definition of a [limit at infinity](/calculus/applications-of-derivatives/how-derivatives-shape-a-graph), with the single change that $n$ ranges over integers. > **Definition (Limit of a sequence).** A sequence $\{a_n\}$ has the limit $L$, > written $\lim_{n\to\infty} a_n = L$ or $a_n \to L$, if for every > $\varepsilon > 0$ there is a corresponding integer $N$ such that > $$ > n > N \quad\Longrightarrow\quad |a_n - L| < \varepsilon. > $$ > If the limit exists the sequence **converges**; otherwise it **diverges**. Geometrically, no matter how narrow the band $(L - \varepsilon, L + \varepsilon)$ is drawn, all terms from some index $N + 1$ onward must fall inside it. A smaller $\varepsilon$ usually forces a larger $N$. If a function of a real variable agrees with the sequence at the integers, its limit at infinity carries over. > **Theorem (Function limits transfer).** If $\lim_{x\to\infty} f(x) = L$ and > $f(n) = a_n$ whenever $n$ is an integer, then $\lim_{n\to\infty} a_n = L$. The converse fails: a sequence can converge while the interpolating function oscillates. The theorem still handles most evaluations, since it lets us reuse everything known about functions. In particular, since $\lim_{x\to\infty} x^{-r} = 0$ for $r > 0$, $$ \lim_{n\to\infty} \frac{1}{n^r} = 0 \qquad (r > 0). $$ If $a_n$ grows without bound we write $\lim_{n\to\infty} a_n = \infty$, meaning that for every $M > 0$ there is an $N$ with $a_n > M$ whenever $n > N$. Such a sequence still diverges; it simply diverges _to infinity_. $$ % caption: Three archetypes. Left: terms settle to a limit $L$ (convergent). % Middle: terms climb past every bound (divergent to $\infty$). Right: terms % jump between two values forever and approach nothing (bounded, divergent). \begin{tikzpicture}[scale=1.0, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} % panel 1: converges \begin{scope} \draw[black, ->] (0,0) -- (3.1,0) node[anchor=north east] {$n$}; \draw[black, ->] (0,0) -- (0,2.7) node[anchor=north east] {$a_n$}; \draw[acc, dashed] (0,2) -- (3,2); \node[anchor=east, acc, font=\scriptsize] at (0,2) {$L$}; \foreach \n/\y in {0.4/1.0, 0.8/1.5, 1.2/1.75, 1.6/1.87, 2.0/1.93, 2.4/1.96, 2.8/1.98} \fill (\n,\y) circle (1.8pt); \node[black!70, anchor=north] at (1.5,-0.15) {converges}; \end{scope} % panel 2: grows without bound \begin{scope}[xshift=4.4cm] \draw[black, ->] (0,0) -- (3.1,0) node[anchor=north east] {$n$}; \draw[black, ->] (0,0) -- (0,2.7) node[anchor=north east] {$a_n$}; \foreach \n/\y in {0.4/0.25, 0.8/0.5, 1.2/0.85, 1.6/1.25, 2.0/1.72, 2.4/2.2, 2.8/2.55} \fill (\n,\y) circle (1.8pt); \node[black!70, anchor=north] at (1.5,-0.15) {grows}; \end{scope} % panel 3: oscillates \begin{scope}[xshift=8.8cm] \draw[black, ->] (0,0) -- (3.1,0) node[anchor=north east] {$n$}; \draw[black, ->] (0,0) -- (0,2.7) node[anchor=north east] {$a_n$}; \draw[black, dashed] (0,1.4) -- (3,1.4); \foreach \n/\y in {0.4/0.55, 0.8/2.25, 1.2/0.55, 1.6/2.25, 2.0/0.55, 2.4/2.25, 2.8/0.55} \fill (\n,\y) circle (1.8pt); \node[black!70, anchor=north] at (1.5,-0.15) {oscillates}; \end{scope} \end{tikzpicture} $$ ## Limit Laws for sequences The [Limit Laws](/calculus/limits-and-continuity/limit-laws-and-the-precise-definition) hold for sequences with the same statements and near-identical proofs. If $\{a_n\}$ and $\{b_n\}$ converge and $c$ is constant, then $$ \lim_{n\to\infty}(a_n \pm b_n) = \lim_{n\to\infty} a_n \pm \lim_{n\to\infty} b_n, \qquad \lim_{n\to\infty}(c\,a_n) = c \lim_{n\to\infty} a_n, $$ $$ \lim_{n\to\infty}(a_n b_n) = \Big(\lim_{n\to\infty} a_n\Big)\Big(\lim_{n\to\infty} b_n\Big), \qquad \lim_{n\to\infty}\frac{a_n}{b_n} = \frac{\lim_{n\to\infty} a_n}{\lim_{n\to\infty} b_n} \ \ \text{if } \lim_{n\to\infty} b_n \ne 0. $$ The **Squeeze Theorem** also transfers. > **Theorem (Squeeze Theorem for sequences).** If $a_n \le b_n \le c_n$ for > $n \ge n_0$ and $\lim_{n\to\infty} a_n = \lim_{n\to\infty} c_n = L$, then > $\lim_{n\to\infty} b_n = L$. A companion fact handles alternating signs: if $\lim_{n\to\infty} |a_n| = 0$, then $\lim_{n\to\infty} a_n = 0$, because $-|a_n| \le a_n \le |a_n|$ squeezes $a_n$ to $0$. Three tactics cover most evaluations: - **Divide by the dominant power.** For $\lim_{n\to\infty} n/(n+1)$, divide top and bottom by $n$: $\;n/(n+1) = 1/(1 + 1/n) \to 1$. - **Push a continuous function inside.** If $a_n \to L$ and $f$ is continuous at $L$, then $f(a_n) \to f(L)$. Since sine is continuous at $0$, $\sin(\pi/n) \to \sin 0 = 0$. - **Squeeze when the terms trap between two known limits.** [l'Hospital's Rule](/calculus/exponential-logarithmic-and-inverse-functions/lhospitals-rule) applies to functions of a real variable, not to a sequence directly; route it through the interpolating function and transfer the result back. > **Worked example.** Evaluate $\lim_{n\to\infty} (\ln n)/n$. > > The sequence has no direct l'Hospital form, so differentiate the related > function $f(x) = (\ln x)/x$, an $\infty/\infty$ form: > > $$ > \lim_{x\to\infty} \frac{\ln x}{x} = \lim_{x\to\infty} \frac{1/x}{1} = 0. > $$ > > Because $f(n) = a_n$ at every integer, the transfer theorem gives > $\lim_{n\to\infty} (\ln n)/n = 0$. The Squeeze Theorem is the only recourse when no real-variable function interpolates the sequence at all. > **Worked example.** Show $\displaystyle\lim_{n\to\infty} \frac{n!}{n^n} = 0$. > > The factorial $x!$ is undefined off the integers, so no function limit is > available. Write the ratio as a product of $n$ fractions and split off the > first: > > $$ > \frac{n!}{n^n} > = \frac{1 \cdot 2 \cdot 3 \cdots n}{n \cdot n \cdot n \cdots n} > = \frac{1}{n}\left(\frac{2 \cdot 3 \cdots n}{n \cdot n \cdots n}\right). > $$ > > Each factor in the parenthesized product is at most $1$, so the product is at > most $1$ and $0 < n!/n^n \le 1/n$. Since $1/n \to 0$, the Squeeze Theorem > forces $n!/n^n \to 0$. ## Powers, oscillation, and divergence Two sequences recur often enough to record. The sequence $\{r^n\}$ behaves according to the size of $r$. $$ % caption: The sequence $r^n$ converges only for $-1 < r \le 1$; outside that % range it grows without bound or oscillates, and $r = -1$ gives the divergent % flip-flop $1, -1, 1, -1, \ldots$ \begin{tikzpicture}[scale=1.0, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \definecolor{red}{HTML}{C0392B} \draw[black, ->] (0,0) -- (5.4,0) node[anchor=north east] {$n$}; \draw[black, ->] (0,-1.4) -- (0,2.6) node[anchor=north east] {$r^n$}; \draw[black, dashed] (0,1) -- (5,1); \node[anchor=east, black] at (0,1) {$1$}; % r>1 growth \foreach \n/\y in {1/1.25, 2/1.56, 3/1.95, 4/2.44} \fill (\n,\y) circle (2pt); \node[anchor=west] at (4.1,2.44) {$r > 1$}; % 0 **Result (Powers of $r$).** The sequence $\{r^n\}$ converges if > $-1 < r \le 1$ and diverges otherwise: > $$ > \lim_{n\to\infty} r^n = > \begin{cases} > 0 & \text{if } -1 < r < 1, \\ > 1 & \text{if } r = 1. > \end{cases} > $$ For $r > 1$ the terms grow to $\infty$; for $r \le -1$ they oscillate or grow in magnitude without settling. The case $r = -1$ gives $\{(-1)^n\} = -1, 1, -1, 1, \ldots$, which visits both $-1$ and $1$ infinitely often and therefore approaches no single number. Oscillation between distinct values is the most common way a bounded sequence diverges. ## Monotonic and bounded sequences Two structural properties let us prove convergence even when the limit is unknown. > **Definition (Monotonic).** A sequence is **increasing** if $a_n < a_{n+1}$ > for all $n \ge 1$, and **decreasing** if $a_n > a_{n+1}$ for all $n \ge 1$. It > is **monotonic** if it is one or the other. > **Definition (Bounded).** A sequence is **bounded above** if there is a number > $M$ with $a_n \le M$ for all $n$, and **bounded below** if there is a number > $m$ with $m \le a_n$ for all $n$. If both hold, it is **bounded**. To show $a_n = 3/(n+5)$ is decreasing, note the denominator grows, so $3/(n+5) > 3/(n+6) = a_{n+1}$. For $a_n = n/(n^2+1)$ the comparison is less obvious, so differentiate $f(x) = x/(x^2+1)$: $f'(x) = (1 - x^2)/(x^2+1)^2 < 0$ for $x > 1$, so $f$ decreases and the sequence decreases too. Neither property alone forces convergence. The bounded sequence $\{(-1)^n\}$ diverges, and the monotonic sequence $a_n = n$ diverges to infinity. Together, however, they are decisive. $$ % caption: An increasing sequence bounded above by $M$: the terms are forced to % crowd together and approach a limit $L \le M$ that they never exceed. \begin{tikzpicture}[scale=1.0, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black, ->] (0,0) -- (8.4,0) node[anchor=north east] {$n$}; \draw[black, ->] (0,0) -- (0,3.9) node[anchor=north east] {$a_n$}; \draw[dashed] (0,3.5) -- (8,3.5); \node[anchor=east] at (0,3.5) {$M$}; \draw[acc, dashed] (0,3.0) -- (8,3.0); \node[anchor=east, acc] at (0,3.0) {$L$}; \definecolor{red}{HTML}{C0392B} \foreach \n/\y in {1/1.3, 2/2.05, 3/2.45, 4/2.68, 5/2.81, 6/2.88, 7/2.93, 8/2.96} \fill (\n,\y) circle (2.4pt); \end{tikzpicture} $$ > **Theorem (Monotonic Sequence Theorem).** Every bounded, monotonic sequence > converges. The proof rests on the **Completeness Axiom** of the real numbers: a nonempty set bounded above has a _least_ upper bound. Suppose $\{a_n\}$ is increasing and bounded above, and let $L$ be the least upper bound of the set $\{a_n : n \ge 1\}$. Given $\varepsilon > 0$, the number $L - \varepsilon$ is not an upper bound (since $L$ is the least one), so $a_N > L - \varepsilon$ for some $N$. Because the sequence increases, $a_n \ge a_N > L - \varepsilon$ for all $n > N$; and $a_n \le L$ always. Hence $0 \le L - a_n < \varepsilon$ for $n > N$ โ€” the definition of $\lim_{n\to\infty} a_n = L$. A symmetric argument, using the greatest lower bound, handles decreasing sequences. The theorem certifies that a limit _exists_ without producing its value. It underlies the convergence arguments for series of positive terms, whose partial sums increase. ## Recursively defined sequences The Monotonic Sequence Theorem is most useful on sequences given by a recurrence, where no formula for $a_n$ is available. Consider $$ a_1 = 2, \qquad a_{n+1} = \tfrac{1}{2}(a_n + 6). $$ Computing terms gives $2, 4, 5, 5.5, 5.75, 5.875, \ldots$, suggesting an increasing sequence approaching $6$. **Increasing.** By induction: $a_2 = 4 > 2 = a_1$. If $a_{k+1} > a_k$, then $a_{k+1} + 6 > a_k + 6$, so $\tfrac{1}{2}(a_{k+1} + 6) > \tfrac{1}{2}(a_k + 6)$, i.e. $a_{k+2} > a_{k+1}$. Thus $a_{n+1} > a_n$ for all $n$. **Bounded above by $6$.** Again by induction: $a_1 = 2 < 6$. If $a_k < 6$, then $a_k + 6 < 12$, so $a_{k+1} = \tfrac{1}{2}(a_k + 6) < 6$. Being increasing and bounded, the sequence converges to some $L$. To find $L$, take the limit of both sides of the recurrence. Since $a_{n+1} \to L$ and $a_n \to L$, $$ L = \tfrac{1}{2}(L + 6) \quad\Longrightarrow\quad 2L = L + 6 \quad\Longrightarrow\quad L = 6. $$ This is the standard two-step method: prove convergence first, then solve a fixed-point equation. Solving $L = \tfrac{1}{2}(L+6)$ is legitimate only once the Monotonic Sequence Theorem guarantees $L$ exists; otherwise the algebra can produce a false answer for a divergent sequence. $$ % caption: Cobweb view of $a_{n+1} = \tfrac{1}{2}(a_n + 6)$: iterating bounces % between the map $y = \tfrac{1}{2}(x+6)$ and the diagonal $y = x$, spiralling % into the fixed point where the two lines cross. \begin{tikzpicture}[scale=0.62, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \definecolor{red}{HTML}{C0392B} \draw[black, ->] (0,0) -- (7.5,0) node[anchor=north east] {$a_n$}; \draw[black, ->] (0,0) -- (0,7.5) node[anchor=north east] {$a_{n+1}$}; \draw[black] (0,0) -- (7.2,7.2) node[anchor=south east, black] {$y = x$}; \draw[acc, very thick] (0,3) -- (7.2,6.6) node[anchor=west] {$y = \tfrac{1}{2}(x+6)$}; % fixed point at (6,6) \fill[red] (6,6) circle (3pt); \node[red, anchor=north west] at (6,6) {$L = 6$}; % cobweb starting at a1=2 \draw[black!70] (2,0) -- (2,4); % up to map \draw[black!70] (2,4) -- (4,4); % across to diagonal \draw[black!70] (4,4) -- (4,5); % up to map \draw[black!70] (4,5) -- (5,5); % across \draw[black!70] (5,5) -- (5,5.5); % up \draw[black!70] (5,5.5) -- (5.5,5.5);% across \node[anchor=north, black!70] at (2,0) {$a_1$}; \end{tikzpicture} $$ A second recurrence shows the method when the map is a square root rather than a line. > **Worked example.** Let $a_1 = \sqrt{2}$ and $a_{n+1} = \sqrt{2 + a_n}$. Show > the sequence converges and find its limit. > > The first terms are $\sqrt 2 \approx 1.414$, $\sqrt{3.414} \approx 1.848$, > $\sqrt{3.848} \approx 1.962$, $\sqrt{3.962} \approx 1.990$, suggesting an > increasing sequence approaching $2$. > > _Bounded above by $2$._ By induction: $a_1 = \sqrt 2 < 2$. If $a_k < 2$, then > $2 + a_k < 4$, so $a_{k+1} = \sqrt{2 + a_k} < \sqrt 4 = 2$. > > _Increasing._ Using $a_k < 2$, the ratio $a_{k+1}/a_k = \sqrt{2 + a_k}/a_k$ > exceeds $1$ because $2 + a_k > a_k^2$ (the quadratic $a_k^2 - a_k - 2 = > (a_k - 2)(a_k + 1)$ is negative for $0 < a_k < 2$). Hence $a_{k+1} > a_k$. > > Being increasing and bounded above, the sequence converges to some $L$. > Taking limits in $a_{n+1} = \sqrt{2 + a_n}$ gives $L = \sqrt{2 + L}$, so > $L^2 - L - 2 = 0$, i.e. $(L-2)(L+1) = 0$. The root $L = -1$ is impossible for a > sequence of positive terms, so $L = 2$. $$ % caption: The recursion $a_{n+1} = \sqrt{2 + a_n}$ starting from $\sqrt 2$: % terms increase and stay below the dashed line $L = 2$, crowding toward it. \begin{tikzpicture}[scale=1.0, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black, ->] (0,0) -- (7.4,0) node[anchor=north east] {$n$}; \draw[black, ->] (0,0) -- (0,2.9) node[anchor=north east] {$a_n$}; \draw[acc, dashed] (0,2.5) -- (7,2.5); \node[anchor=east, acc] at (0,2.5) {$L = 2$}; \foreach \n/\y in {1/1.768, 2/2.310, 3/2.452, 4/2.488, 5/2.497, 6/2.499, 7/2.500} \fill (\n,\y) circle (2.2pt); \end{tikzpicture} $$ ## Summary of sequence facts | Idea | Statement | | --- | --- | | Limit of a sequence | $a_n \to L$: terms stay within $\varepsilon$ of $L$ past some $N$ | | Powers $r^n$ | converge iff $-1 < r \le 1$ (to $0$, or to $1$ at $r=1$) | | Squeeze Theorem | $a_n \le b_n \le c_n$ with $a_n, c_n \to L$ forces $b_n \to L$ | | Monotonic Sequence Theorem | bounded $+$ monotonic $\Rightarrow$ convergent | | Two-step recursion method | prove convergence, then solve the fixed-point equation $L = g(L)$ | The sum of an infinite series is defined as the limit of its sequence of partial sums, so every result above transfers directly to [series](/calculus/sequences-and-series/series-and-the-integral-test).