The sequence \(b_n\) is the Balmer series of absorption wavelengths of the hydrogen atom in nanometers. It plays a key role in spectroscopy.

EXAMPLE 1 Recursive Sequence

Compute \(a_2, a_3, a_4\) for the sequence defined recursively by \[ a_1 = 1,\qquad a_n = \frac12\left(a_{n-1}+\frac2{a_{n-1}}\right) \]

Solution

\[ \begin{alignat*}{2} a_2 = \frac12\left(a_1+\frac2{a_1}\right)~= \frac12\left(1+\frac2{1}\right) &~= \frac32 = 1.5\\ a_3 = \frac12\left(a_2+\frac2{a_2}\right) = \frac12\left(\frac32+\frac2{3/2}\right)~ &= \frac{17}{12}\approx 1.4167\\ a_4 =\; \frac12\left(a_3+\frac2{a_3}\right) = \frac12\left(\frac{17}{12}+\frac2{17/12}\right)~ &= \frac{577}{408}\approx 1.414216 \end{alignat*} \]

DEFINITION Limit of a Sequence

We say that \(\{a_n\}\) converges to a limit \(L\), and we write \[ \lim_{n\to\infty} a_n = L\qquad\textrm{or}\qquad a_n\to L \] if, for every \(\epsilon >0\), there is a number \(M\) such that \(|a_n-L|<\epsilon\) for all \(n > M\).

• If no limit exists, we say that \(\{a_n\}\) diverges.
• If the terms increase without bound, we say that \(\{a_n\}\) diverges to infinity.

EXAMPLE 2 Proving Convergence

Let \(a_n\! =\! \frac{n+4}{n+1}\). Prove formally that \( \displaystyle\lim_{n\to\infty} a_n\! =\! 1\).

Solution The definition requires us to find, for every \(\epsilon>0\), a number \(M\) such that \[ \begin{equation*} |a_n - 1|<\epsilon \qquad\textrm{for all \(n>M\)}\tag{1} \end{equation*} \]

We have \[ |a_n-1|~=\Bigg|\frac{n+4}{n+1} - 1\Bigg|~= \frac3{n+1} \] Therefore, \(|a_n-1|<\epsilon\) if \[ \frac3{n+1} < \epsilon \qquad\textrm{or}\qquad n >\frac{3}{\epsilon}-1 \] In other words, Eq. (1) is valid with \(M = \tfrac{3}{\epsilon}-1\). This proves that \(\lim\limits_{n\to\infty} a_n = 1\).

THEOREM 1 Sequence Defined by a Function

If \(\lim\limits_{x\to\infty}f(x)\) exists, then the sequence \(a_n = f(n)\) converges to the same limit: \[ \lim\limits_{n\to\infty} a_n = \lim\limits_{x\to\infty}f(x) \]

EXAMPLE 3

Find the limit of the sequence \[ \dfrac{2^2-2}{2^2},\quad \dfrac{3^2-2}{3^2},\quad \dfrac{4^2-2}{4^2},\quad \dfrac{5^2-2}{5^2},\quad \dots \]

Solution This is the sequence with general term \[ a_n=\frac{n^2-2}{n^2} = 1 - \frac2{n^2} \] Therefore, we apply Theorem 1 with \(f(x)=1-\tfrac{2}{x^2}\): \[ \lim\limits_{n\to\infty} a_n =\lim\limits_{x\to\infty} \left(1-\frac2{x^2}\right)=1-\lim\limits_{x\to\infty}\frac2{x^2}=1 - 0 = 1 \]

EXAMPLE 4

Calculate \(\lim\limits_{n\to\infty} \frac{n+\ln n}{n^2}\).

Solution Apply Theorem 1, using L'Hôpital’s Rule in the second step: \[ \lim\limits_{n\to\infty}\frac{n+\ln n}{n^2} = \lim\limits_{x\to\infty}\frac{x+\ln x}{x^2} = \lim\limits_{x\to\infty}\frac{1+(1/x)}{2x} = 0 \]

EXAMPLE 5 Balmer Wavelengths

Calculate the limit of the Balmer wavelengths \(b_n= \frac{364.5n^2}{n^2-4}\), where \(n \geq 3\).

Solution Apply Theorem 1 with \(f(x)= \frac{364.5x^2}{x^2-4}\): \[ \lim\limits_{n\to\infty}b_n = \lim\limits_{x\to\infty}\frac{364.5x^2}{x^2-4} = \lim\limits_{x\to\infty}\frac{364.5}{1-4/x^2} =\dfrac{364.5}{\lim\limits_{x\to\infty}(1-4/x^2)} = 364.5 \]

EXAMPLE 6 Geometric Sequences with \(r \geq 0\)

Prove that for \(r\ge 0\) and \(c>0\), \[\begin{align*} \lim\limits_{n\to\infty}cr^n = \begin{cases} 0 & \text{if}\quad 0\le r <1\\ c & \text{if}\quad r=1\\ \infty & \text{if}\quad r > 1 \end{cases} \end {align*}\]

Solution Set \(f(r)=cr^x\). If \(0\le r<1\), then (Figure 10.9) \[ \lim\limits_{n\to\infty}cr^n = \lim\limits_{x\to\infty}f(x) = c\lim\limits_{x\to\infty}r^x = 0 \]

If \(r>1\), then both \(f(x)\) and the sequence \(\{cr^n\}\) diverge to \(\infty\) (because \(c>0\)) (Figure 10.8).

If \(r=1\), then \(cr^n=c\) for all \(n\), and the limit is \(c\).

THEOREM 2 Limit Laws for Sequences

Assume that \(\{a_n\}\) and \(\{b_n\}\) are convergent sequences with \[ \lim\limits_{n\to\infty}a_n = L,\qquad \lim\limits_{n\to\infty}b_n = M \]

Then:

1. \(\lim\limits_{n\to\infty}(a_n\pm b_n) = \lim\limits_{n\to\infty} a_n \pm\lim\limits_{n\to\infty} b_n = L\pm M\)
2. \(\lim\limits_{n\to\infty} a_nb_n = \Big(\lim\limits_{n\to\infty}a_n\Big)\Big(\lim\limits_{n\to\infty}b_n\Big) = LM\)
3. \(\lim\limits_{n\to\infty}\frac{a_n}{b_n} = \frac{\lim\limits_{n\to\infty}a_n}{ \lim\limits_{n\to\infty}b_n} = \frac{L}M \textrm{if}~M \ne 0\)
4. \(\lim\limits_{n\to\infty} ca_n = c\lim\limits_{n\to\infty}a_n = cL\) for any constant \(c\)

THEOREM 3 Squeeze Theorem for Sequences

Let \(\{a_n\}\), \(\{b_n\}\), \(\{c_n\}\) be sequences such that for some number \(M\), \[ b_n\le a_n\le c_n\quad\textrm{for \(n>M\)}\qquad\textrm{and}\qquad \lim\limits_{n\to\infty}b_n = \lim\limits_{n\to\infty}c_n = L \] Then \(\lim\limits_{n\to\infty}a_n = L\).

Reminder

\(n!\)(\(n\)-factorial) is the number \[ n! = n(n-1)(n-2)\cdots 2\cdot 1 \] For example, \(4!=4\cdot 3\cdot 2\cdot 1 = 24\).

EXAMPLE 7

Show that if \(\lim\limits_{n\to\infty}|a_n|=0\), then \(\lim\limits_{n\to\infty}a_n=0\).

Solution We have \[ -|a_n|\le a_n \le |a_n| \] By hypothesis, \(\lim\limits_{n\to\infty} |a_n| = 0\), and thus also \(\lim\limits_{n\to\infty}-|a_n| =-\lim\limits_{n\to\infty}|a_n|=0\). Therefore, we can apply the Squeeze Theorem to conclude that \(\lim\limits_{n\to\infty}a_n=0\).

EXAMPLE 8 Geometric Sequences with \(r<0\)

Prove that for \(c\ne 0\), \[ \lim\limits_{n\to\infty}cr^n = \begin{cases} 0 & \text{if}\quad -1< r < 0\\ \textrm{diverges} & \text{if} \quad r \le -1 \end{cases} \]

Solution If \(-1< r < 0\), then \(0<|r|<1\) and \(\lim\limits_{n\to\infty}|cr^n|=0\) by Example 6. Thus \(\lim\limits_{n\to\infty}cr^n=0\) by Example 7. If \(r= -1\), then the sequence \(cr^{n} = (-1)^nc\) alternates in sign and does not approach a limit. The sequence also diverges if \(r<-1\) because \(cr^n\) alternates in sign and \(|cr^{n}|\) grows arbitrarily large.

EXAMPLE 9

Prove that \(\lim\limits_{n\to\infty}\frac{R^n}{n!} = 0\) for all \(R\).

Solution First notice that if \(-1 \le R \le 1\), then \(-1 \le R^n \le 1\) also for all \(n\) and thus the numerator is bounded. Since the denominator diverges to \(\infty\) as \(n \longrightarrow \infty\), this implies that \(\lim\limits_{n\to\infty}\frac{R^n}{n!} = 0\) for all \(R\) where \(-1 \le R \le 1\). Assume next that \(R>1\) and let \(M\) be the positive integer such that \[ M\le R < M+1 \] For \(n>M\), we write \(R^n/n!\) as a product of \(n\) factors: \[ \begin{equation*} \frac{R^n}{n!} = \underbrace{\left(\frac{R}{1}\frac{R}{2}\cdots\frac{R}{M}\right)} _{\textrm{Call this constant \(C\)}} \underbrace{\left(\frac{R}{M+1}\right)\left(\frac{R}{M+2}\right) \cdots\left(\frac{R}{n}\right)}_{\textrm{Each factor is less than 1}} \le C\left(\frac{R}{n}\right)\tag{2} \end{equation*} \] The first \(M\) factors are \(\ge 1\) and the last \(n-M\) factors are \(<1\). If we lump together the first \(M\) factors and call the product \(C\), and drop all the remaining factors except the last factor \(R/n\), we see that \[ 0\le \frac{R^n}{n!}\le \frac{CR}{n} \] Since \({CR}/{n}\to 0\), the Squeeze Theorem gives us \(\lim\limits_{n\to\infty}{R^n}/{n!} = 0\) as claimed. If \(R<-1\), the limit is also zero by Example 7 because \(\left|{R^n}/{n!}\right|\) tends to zero.

THEOREM 4

If \(f(x)\) is continuous and \(\lim\limits_{n\to\infty} a_n = L\), then \[ \lim\limits_{n\to\infty}f(a_n) = f\Bigl(\lim\limits_{n\to\infty}a_n\Bigr) = f(L) \] In other words, we may “bring a limit inside a continuous function.”

EXAMPLE 10

Apply Theorem 4 to the sequence \(a_n = \frac{3n}{n+1}\) and to the functions

(a) \(f(x) = e^x\) and (b) \(g(x) = x^2\).

Solution Observe first that \[ L = \lim\limits_{n\to\infty}a_n = \lim\limits_{n\to\infty} \frac{3n}{n+1} = \lim\limits_{n\to\infty} \frac{3}{1+n^{-1}} = 3 \]

1. With \(f(x)=e^x\) we have \( f(a_n) = e^{a_n} = e^{\frac{3n}{n+1}}\). According to Theorem 4, \[ \lim\limits_{n\to\infty} e^{\frac{3n}{n+1}} =\lim\limits_{n\to\infty} f(a_n) = f\Bigl(\lim\limits_{n\to\infty} a_n\Bigr) = e^{\mkern1mu\lim\limits_{n\to\infty} \frac{3n}{n+1}} = e^3 \]
2. With \(g(x)=x^2\) we have \(g(a_n)=a_n^2\), and according to Theorem 4, \[ \lim\limits_{n\to\infty} \left(\frac{3n}{n+1}\right)^2 =\lim\limits_{n\to\infty} g(a_n) = g\Bigl(\lim\limits_{n\to\infty} a_n\Bigr) = \left(\lim\limits_{n\to\infty} \frac{3n}{n+1}\right)^2 = 3^2 = 9 \]

DEFINITION Bounded Sequences

A sequence \(\{a_n\}\) is:

• Bounded from above if there is a number \(M\) such that \(a_n\le M\) for all \(n\). The number \(M\) is called an upper bound.
• Bounded from below if there is a number \(m\) such that \(a_n\ge m\) for all \(n\). The number \(m\) is called a lower bound.

The sequence \(\{a_n\}\) is called bounded if it is bounded from above and below. A sequence that is not bounded is called an unbounded sequence.

THEOREM 5 Convergent Sequences Are Bounded

If \(\{a_n\}\) converges, then \(\{a_n\}\) is bounded.

Proof

Let \(L=\lim\limits_{n\to\infty}a_n\). Then there exists \(N>0\) such that \(|a_n - L|<1\) for \(n>N\). In other words, \[ L-1<a_n<L+1\qquad\textrm{for \(n>N\)} \] If \(M\) is any number larger than \(L+1\) and also larger than the numbers \(a_1, a_2, \dots, a_N\), then \(a_n<M\) for all \(n\). Thus, \(M\) is an upper bound. Similarly, any number \(m\) smaller than \(L-1\) and also smaller than the numbers \(a_1, a_2, \dots, a_N\) is a lower bound.

THEOREM 6 Bounded Monotonic Sequences Converge

• If \(\{a_n\}\) is increasing and \(a_n\le M\), then \(\{a_n\}\) converges and \(\lim\limits_{n\to\infty} a_n \le M\).
• If \(\{a_n\}\) is decreasing and \(a_n\ge m\), then \(\{a_n\}\) converges and \(\lim\limits_{n\to\infty} a_n \ge m\).

EXAMPLE 11

Verify that \(a_n=\sqrt{n+1}-\sqrt n\) is decreasing and bounded below. Does \(\lim\limits_{n\to\infty}a_n\) exist?

Solution The function \(f(x)=\sqrt{x+1}-\sqrt x\) is decreasing because its derivative is negative: \[ f'(x) = \frac1{2\sqrt{x+1}} - \frac1{2\sqrt{x}} < 0 \qquad\textrm{for \(x>0\)} \] It follows that \(a_n = f(n)\) is decreasing (see Table ). Furthermore, \(a_n > 0\) for all \(n\), so the sequence has lower bound \(m=0\). Theorem 6 guarantees that \(L=\lim\limits_{n\to\infty}a_n\) exists and \(L\ge 0\). In fact, we can show that \(L=0\) by noting that \(f(x) = 1/(\sqrt{x+1}+\sqrt x)\) and hence \(\lim\limits_{x\to\infty}f(x)=0\).

EXAMPLE 12

Show that the following sequence is bounded and increasing: \[ a_1=\sqrt{2},\quad a_2=\sqrt{2\sqrt{2}},\quad a_3=\sqrt{2\sqrt{2\sqrt{2}}},\quad \dots\quad \] Then prove that \(L=\lim\limits_{n\to\infty}a_n\) exists and compute its value.

Solution If we knew in advance that the limit \(L\) exists, we could find its value as follows. The idea is that \(L\) “contains a copy” of itself under the square root sign: \[ L = \sqrt{2\sqrt{2\sqrt{2\sqrt{2\cdots }}}} = \sqrt{2\left(\sqrt{2\sqrt{2\sqrt{2\cdots }}}\right)} = \sqrt{2L} \] Thus \(L^2=2L\), which implies that \(L = 2\) or \(L=0\). We eliminate \(L=0\) because the terms \(a_n\) are positive and increasing (as shown below), so we must have \(L=2\) (see Table ).

This argument is phrased more formally by noting that the sequence is defined recursively by \[ a_1=\sqrt{2},\qquad a_{n+1}=\sqrt{2a_n} \] If \(a_n\) converges to \(L\), then the sequence \(b_n = a_{n+1}\) also converges to \(L\) (because it is the same sequence, with terms shifted one to the left). Then, using Theorem 4, we would have \[ L = \lim\limits_{n\to\infty} a_{n+1} = \lim\limits_{n\to\infty} \sqrt{2a_n} =\sqrt{2\lim\limits_{n\to\infty}a_n}=\sqrt{2L} \] However, none of this is valid unless we know in advance that the limit \(L\) exists. By Theorem 6, it suffices to show that \(\{a_n\}\) is bounded above and increasing.

Step 1. Show that \(\{a_n\}\) is bounded above.

We claim that \(M=2\) is an upper bound. We certainly have \(a_1 < 2\) because \(a_1=\sqrt{2}\approx 1.414\). On the other hand, \[ \begin{equation*} {if} a_n<2,\qquad {then}\qquad a_{n+1} < 2\tag{3} \end{equation*} \]

This is true because \(a_{n+1}=\sqrt{2a_n}<\sqrt{2\cdot 2}=2\). Now, since \(a_1 < 2\), we can apply (3) to conclude that \(a_2<2\). Similarly, \(a_2<2\) implies \(a_3<2\), and so on for all \(n\). Formally speaking, this is a proof by induction.

Step 2. Show that \(\{a_n\}\) is increasing

Since \(a_n\) is positive and \(a_n < 2\), we have \[ a_{n+1} = \sqrt{2a_n}> \sqrt{a_n\cdot a_n} = a_n \] This shows that \(\{a_n\}\) is increasing.