Limits of Sequences and Cauchy’s Criterion

A sequence of numbers is really a function which assigns to each positive integer a number. We denote it customarily by ${a_{n}}$ , where $a_{n} = f (n)$ for each integer $n > 0$ . Examples are:
a) $1, - 1, 1, - 1, 1, - 1, \dots;$
$a_{n} = 1$ if $n$ is odd, $- 1$ if $n$ is even.
b) $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots;$
$a_{n} = \frac{1}{n} \cdot$
c) $1, 1 + 1, 1 + 1 + \frac{1}{2!}, \dots;$
$a_{n} = 1 + \frac{1}{1!} + \frac{1}{2!} + \dots + \frac{1}{(n - 1)!} \cdot$

The sequence ${a_{n}}$ is said to have the limit $a$ if for every given positive number $ϵ$ (usually considered as small), we can find an integer $N$ such that whenever $n > N$ , we have $a - ϵ < a_{n} < a + ϵ$ , or, what is the same, $| a - a_{n} | < ϵ$ . That is to say, I can only assert that the limit is $a$ if every opponent who chooses to challenge me on the matter can be dealt with; I must be prepared to give a procedure for finding an $N$ with the desired property on the basis of any $ϵ$ with which he may wish to confront me. If ${a_{n}}$ has the limit $a$ , we write either

lim_{n \to \infty} a_{n} = a, or simply lim a_{n} = a \cdot

The former is read, "The limit as $n$ tends to infinity of $a_{n}$ is $a$ ". Note that we have nowhere defined any object called "infinity", but rather we have chosen this whole idiom to refer to the behavior which has been given a precise definition above.

The sequence a) has no limit (try to meet the challenge of $ϵ = \frac{1}{2}$ ). The sequence b) has the limit $0$ , and the sequence c) has the limit $e$ , as we have seen in Section 17. Suppose however that we are given a sequence ${a_{n}}$ and it is not evident what its limit is. Can we tell whether it has a limit without first guessing the limit? The answer to this is yes, by means of several possible devices. The most powerful one, the Cauchy convergence criterion, will be given with its proof as an appendix to this section. One which is somewhat simpler may be applied if the sequence is increasing: $a_{1} \leq a_{2} \leq a_{3} \leq \dots \cdot$

Theorem 1

Let ${a_{n}}$ be increasing, and let ${a_{n}}$ have an upper bound. Then $lim a_{n} = a$ for some number $a$ .

Let $a$ be the least upper bound for ${a_{n}}$ . Let $ϵ > 0$ . We must show that there is an $N$ such that whenever $n > N, a - ϵ < a_{n} < a + ϵ$ . But $a_{n} < a + ϵ$ for **all** $n$ , since $a_{n} \leq a$ for all $n$ ( $a$ is an upper bound). Now $a - ϵ$ **is not** an upper bound, so there is an $N$ such that $a_{N} > a - ϵ$ . If $n > N$ , then $a_{n} \geq a_{N} > a - ϵ$ . Thus for $n > N$ , we have $a - ϵ < a_{n} < a + ϵ$ , which was to be proved. (Note that $N$ will in general depend on $ϵ$ .)

If we did not know that the limit of the sequence c) is $e$ , we could show by this theorem that it has a limit. For the sequence is obviously increasing, and

\begin{aligned} 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{(n-1)!} &\le 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{n-2}} \\ &= 1 + \frac{1 - (\frac{1}{2})^{n-1}}{1 - \frac{1}{2}} < 3 \quad \text{for all } n \cdot \end{aligned}

Therefore we have a bounded increasing sequence. The theorem can also be applied, using negatives, to show that if a decreasing sequence has a lower bound, it has a limit. (Note also that two numbers $a$ and $b$ cannot both be limits of ${a_{n}}$ ; for then we could not meet the challenge of $ϵ = \frac{1}{2} | b - a |$ for either point.)

As an example of a sequence whose limit we can not guess, let $a_{n} = 1 + \frac{1}{2} + \dots + \frac{1}{n} - \log n$ . Then

a_{n + 1} - a_{n} = \frac{1}{n + 1} - \log (n + 1) + \log n = \frac{1}{n + 1} - \log (1 + \frac{1}{n}) \cdot

But if we recall from Section 11 the inequality

\log (1 + \frac{1}{n}) \geq \frac{1}{n + 1}, we see that a_{n} \geq a_{n + 1},

or that the sequence is decreasing. Now let us show that it has $0$ as a lower bound. To do this, recall the other side of the same inequality on $\log x$ :

\frac{1}{n} \geq \log (1 + \frac{1}{n}) = \log (n + 1) - \log n \cdot

Thus

\begin{aligned} a_n &\ge (\log 2 - \log 1) + (\log 3 - \log 2) + \dots + (\log(n+1) - \log n) - \log n \\ &= (\log 2 - \log 2) + (\log 3 - \log 3) + \dots + (\log n - \log n) + \log(n+1) - \log n \\ &= \log(n+1) - \log n \ge 0 \quad \text{for all } n \cdot \end{aligned}

Therefore

lim_{n \to \infty} (1 + \frac{1}{2} + \dots + \frac{1}{n} - \log n) = C

exists, and is in fact non-negative. Its first few decimals are $.5772 \dots$ . It is called Euler's constant, and it is not even known whether it is rational.

Next we prove some properties of limits of sequences, including the linearity of the limits. Some others are left as exercises. In general, let ${a_{n}}$ be a sequence with limit $a$ , ${b_{n}}$ a sequence with limit $b$ , and $c$ a constant.

Theorem 2

lim (a_{n} + b_{n}) = a + b \cdot

Let $ϵ > 0$ be given. Then we can find $N_{1}$ such that for all $n > N_{1}, - ϵ < a_{n} - a < ϵ$ , and $N_{2}$ such that for all $n > N_{2}, - ϵ < b_{n} - b < ϵ$ . Let $N$ be the **larger of $N_{1}$ and $N_{2}$ **, written $N = Max (N_{1}, N_{2})$ . Then for all $n > N, (a_{n} + b_{n}) - (a + b)$ has the property that

- 2 ϵ < (a_{n} + b_{n}) - (a + b) < 2 ϵ \cdot

The factor of $2$ makes no difference, since we could have started with $\frac{ϵ}{2}$ instead of $ϵ$ and arrived at $- ϵ < (a_{n} + b_{n}) - (a + b) < ϵ$ for all n > N' for a suitable N'. Therefore the theorem is proved.

Theorem 3

lim (c a_{n}) = c a \cdot

If $c = 0$ , then all $c a_{n} = c a = 0$ , and for any given $ϵ > 0$ , we can take $N = 1$ . Then $- ϵ < c a_{n} - c a = 0 < ϵ$ , for all $n > N$ . Now let $c \neq 0$ , and let $ϵ > 0$ be given. Let $N_{1}$ be such that $- ϵ < a_{n} - a < ϵ$ for all $n > N_{1}$ , or in other words, $| a_{n} - a | < ϵ$ for $n > N_{1}$ . Then

| c a_{n} - c a | = | c | | a_{n} - a | < | c | ϵ for n > N_{1} \cdot

We need only start with $\frac{ϵ}{| c |}$ instead of $ϵ$ to get a result of the form

| c a_{n} - c a | < ϵ for n > N \cdot

Theorem 4

lim (a_{n} b_{n}) = a b \cdot

PROOF IN PART.

\begin{aligned} |a_n b_n - ab| &= |a_n b_n - a_n b + a_n b - ab| \\ &\le |a_n b_n - a_n b| + |a_n b - ab| = |a_n| |b_n - b| + |a_n - a| |b| \cdot \end{aligned}

Now consider ${a_{n}}$ , with $ϵ = 1$ . There is an $N_{1}$ such that for $n > N_{1}$ , $| a_{n} - a | < 1$ . Now we claim that $| a_{n} | - | a | \leq | a_{n} - a |$ . For

| a_{n} | = | (a_{n} - a) + a | \leq | a_{n} - a | + | a |,

| a_{n} | - | a | \leq | a_{n} - a | \cdot

Thus we have $| a_{n} | - | a | \leq | a_{n} - a | < 1$ , or $| a_{n} | < 1 + | a |$ for all $n > N_{1}$ . Therefore if $n > N_{1}$ ,

| a_{n} b_{n} - a b | \leq (| a | + 1) | b_{n} - b | + | a_{n} - a | | b | \cdot

The remainder of the proof is left as an exercise.

Exercise 1.

Suppose $lim a_{n} = a, a \neq 0$ .
Show that there is a positive number $r$ and an integer $N_{1}$ such that for all $n > N_{1}, | a_{n} | > r$ .

Now show that $lim (\frac{1}{a_{n}}) = \frac{1}{a}$ , in the sense that for any given $ϵ > 0$ there is an $N$ such that for all $n > N, a_{n} \neq 0$ and $| \frac{1}{a_{n}} - \frac{1}{a} | < ϵ$ .

Next let $f (x)$ be a function defined on an interval containing a point $x_{0}$ , with the possible exception that $f (x_{0})$ may not be defined. We say that $a$ is the limit as $x$ approaches $x_{0}$ of $f (x)$ , written

lim_{x \to x_{0}} f (x) = a,

if for every given $ϵ > 0$ we can find a $δ > 0$ such that for every $x$ in the interval $0 < | x - x_{0} | < δ$ we have $| f (x) - a | < ϵ$ .

Exercise 2.

Let

lim_{x \to x_{0}} f (x) = a, lim_{x \to x_{0}} g (x) = b,

$c$ a constant. Prove:

$lim_{x \to x_{0}} (f (x) + g (x)) = a + b$ .
$lim_{x \to x_{0}} (c f (x)) = c a$ .
$lim_{x \to x_{0}} (f (x) g (x)) = a b$ .
If $a \neq 0, lim_{x \to x_{0}} (\frac{1}{f (x)}) = \frac{1}{a} \cdot$
$lim_{x \to x_{0}} f (x) = a$ means the same as $lim_{h \to 0} f (x_{0} + h) = a$ .

Appendix: Cauchy's Criterion

As a variation, we prove this theorem for the limit of a function, and leave its proof for sequences as an exercise.

Theorem 5

Let $f (x)$ be defined on an interval containing $x_{0}$ , but not necessarily at $x_{0}$ . Then

lim_{x \to x_{0}} f (x)

exists if and only if $f (x)$ satisfies the following condition: For every $ϵ > 0$ there is a $δ > 0$ such that if $x_{1}$ and $x_{2}$ are any numbers with $0 < | x_{1} - x_{0} | < δ, 0 < | x_{2} - x_{0} | < δ$ , then $| f (x_{1}) - f (x_{2}) | < ϵ \cdot$

First suppose the limit exists, say

lim_{x \to x_{0}} f (x) = a \cdot

Let $ϵ > 0$ be given. Then there is a $δ > 0$ such that if $0 < | x - x_{0} | < δ$ , we have $| f (x) - a | < \frac{ϵ}{2} \cdot$ Let $0 < | x_{1} - x_{0} | < δ, 0 < | x_{2} - x_{0} | < δ$ . Then

\begin{aligned} |f(x_1) - f(x_2)| &= |f(x_1) - a + a - f(x_2)| \\ &\le |f(x_1) - a| + |f(x_2) - a| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \cdot \end{aligned}

Thus the theorem is proved in one direction: if the limit exists, the condition is satisfied.

Now suppose the condition is satisfied. First take $ϵ = 1$ ; then let $0 < | x_{1} - x_{0} | < δ_{0}$ , where $δ_{0}$ satisfies the condition for $ϵ = 1$ . Now if $x$ is any number such that $0 < | x - x_{0} | < δ_{0}$ , we have $| f (x) - f (x_{1}) | < 1$ , or $| f (x) | < | f (x_{1}) | + 1$ , i.e., $f (x)$ is bounded from above and from below for $0 < | x - x_{0} | < δ_{0}$ .

Next let $δ_{1} = Min (1, δ_{0}), δ_{2} = Min (\frac{1}{2}, δ_{0}), \dots, δ_{n} = Min (\frac{1}{n}, δ_{0})$ . (From some $n$ on, all $δ_{n} = \frac{1}{n}$ , since eventually $\frac{1}{n} < δ_{0} \cdot$ ) On each of the intervals $I_{n} : 0 < | x - x_{0} | < δ_{n}, f (x)$ is bounded from above and from below. Let $a_{n}$ be the least upper bound of the values of $f (x)$ on $I_{n}$ . The sequence ${a_{n}}$ is decreasing, since a least upper bound on $I_{n}$ is an upper bound on $I_{n + 1}$ . Moreover, all $a_{n} \geq L$ , where $L$ is any lower bound for $f (x)$ on $I_{1}$ . Therefore $lim a_{n}$ exists. Let us call it $a$ . Now we shall prove that

a = lim_{x \to x_{0}} f (x) \cdot

Let $ϵ > 0$ be given. Then there is an $N$ such that for all $n > N, | a - a_{n} | < \frac{ϵ}{3} \cdot$ Since $a_{n}$ is the least upper bound for $f (x)$ in $I_{n}$ , there is an $x_{n}$ in $I_{n}$ , such that $| a_{n} - f (x_{n}) | < \frac{ϵ}{3}$ , and this for each $n$ . There is also a \delta' such that whenever 0 < |x - x_0| < \delta', 0 < |x' - x_0| < \delta', then |f(x) - f(x')| < \frac{\epsilon}{3} \cdot Now take $n > N$ such that \delta_n < \delta' \cdot We assert that $δ_{n}$ is a suitable $δ$ for the definition of limit. That is, let $0 < | x - x_{0} | < δ_{n} \cdot$ Then 0 < |x - x_0| < \delta', and also 0 < |x_n - x_0| < \delta_n < \delta' \cdot Therefore

| f (x) - f (x_{n}) | < \frac{ϵ}{3} \cdot

Now we know that $| f (x_{n}) - a_{n} | < \frac{ϵ}{3} \cdot$ and since $n > N$ ,

| a_{n} - a | < \frac{ϵ}{3} \cdot

Therefore

\begin{aligned} |f(x) - a| &= |f(x) - f(x_n) + f(x_n) - a_n + a_n - a| \\ &\le |f(x) - f(x_n)| + |f(x_n) - a_n| + |a_n - a| \\ &< \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \cdot \end{aligned}

We see that whenever $0 < | x - x_{0} | < δ_{n}$ , we have $| f (x) - a | < ϵ$ . Therefore

lim_{x \to x_{0}} f (x) = a,

and the proof is complete.

Exercise 3.

Prove Cauchy's criterion for sequences; $lim a_{n}$ exists if and only if the following condition holds: For any $ϵ > 0$ there is an $N$ such that whenever $m > N$ and $n > N$ , then $| a_{m} - a_{n} | < ϵ$ .

Sequences and Limits

Appendix: Cauchy's Criterion

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