Sequence Theorems, calculus lesson

Checking every sequence limit straight from the definition would be slow and tiring. In practice you break a sequence into pieces whose limits you already know, trap it between two simpler sequences, or relate it to a function you can handle with ordinary calculus. The theorems below are what make those moves legal.

Key terms

Uniqueness of limits

Definition A sequence $(a_n)$ converges to $L$ if for every $\varepsilon > 0$ there is an integer $N$ such that $\lvert a_n - L \rvert < \varepsilon$ for all $n \ge N$ . If $(a_n)$ converges to $L$ and also to $M$ , then $L = M$ .

In plain words A sequence cannot settle down near two different numbers at once. So it makes sense to speak of the limit and write $\lim_{n \to \infty} a_n = L$ .

Example Suppose $a_n \to L$ and $a_n \to M$ with $L \ne M$ , and take half their distance as the tolerance,

\varepsilon = \frac{\lvert L - M \rvert}{2} > 0.

Far enough out, $a_n$ is within $\varepsilon$ of both, so

\lvert L - M \rvert \le \lvert L - a_n \rvert + \lvert a_n - M \rvert < 2\varepsilon = \lvert L - M \rvert,

which is impossible. This is also why $a_n = (-1)^n$ has no limit. Its terms keep landing on both $1$ and $-1$ , which are $2$ apart, so they cannot all stay within $\frac{1}{2}$ of a single number.

Limit laws

Definition Suppose $a_n \to L$ and $b_n \to M$ , and let $c$ be a constant. Then

a_n \pm b_n \to L \pm M, \qquad c\,a_n \to cL, \qquad a_n b_n \to LM,

and if $M \ne 0$ and $b_n \ne 0$ for every $n$ , then

\frac{a_n}{b_n} \to \frac{L}{M}.

Also $a_n^{\,p} \to L^p$ for every positive integer $p$ .

In plain words Limits of sequences respect arithmetic, as long as every piece has a finite limit and you never divide by a limit of zero.

Example Divide the top and bottom by the highest power of $n$ .

\frac{3n^2 + 1}{2n^2 - n} = \frac{3 + \frac{1}{n^2}}{2 - \frac{1}{n}} \to \frac{3 + 0}{2 - 0} = \frac{3}{2}.

The denominator $2 - \frac{1}{n}$ is never zero for $n \ge 1$ and has limit $2 \ne 0$ , so the quotient law applies.

Comparison of limits

Definition If $a_n \to L$ , $b_n \to M$ and $a_n \le b_n$ for all $n \ge N$ for some $N$ , then $L \le M$ .

In plain words An inequality between terms survives in the limit, but only as $\le$ . Strict inequalities between terms can turn into equality.

Example For every $n \ge 1$ we have $0 < \frac{1}{n}$ , yet $\lim \frac{1}{n} = 0$ . So the strict inequality $0 < \frac{1}{n}$ only gives $0 \le \lim \frac{1}{n}$ , and here equality happens.

Squeeze theorem

Definition If $a_n \le b_n \le c_n$ for all $n \ge N$ and

\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = L,

then $\lim_{n \to \infty} b_n = L$ . A useful special case is that $\lvert b_n \rvert \to 0$ implies $b_n \to 0$ .

In plain words Trap the hard sequence between two easy ones that head to the same place. It has no room to go anywhere else.

Example Since $-1 \le \sin n \le 1$ , for $n \ge 1$

-\frac{1}{n} \le \frac{\sin n}{n} \le \frac{1}{n}.

Both outer sequences go to $0$ , so $\frac{\sin n}{n} \to 0$ . Notice that $\sin n$ itself has no limit. The squeeze works anyway.

Continuous functions of sequences

Definition If $a_n \to L$ , every $a_n$ lies in the domain of $f$ , and $f$ is continuous at $L$ , then

\lim_{n \to \infty} f(a_n) = f\!\left(\lim_{n \to \infty} a_n\right) = f(L).

In plain words A continuous function lets you pass the limit inside. Find where the inside goes, then apply $f$ .

Example The inside sequence $\frac{4n + 1}{n} = 4 + \frac{1}{n}$ tends to $4$ , and $\sqrt{x}$ is continuous at $4$ , so

\sqrt{\frac{4n + 1}{n}} \to \sqrt{4} = 2.

In the same way $e^{1/n} \to e^0 = 1$ .

Subsequences

Definition A subsequence of $(a_n)$ is a sequence $(a_{n_k})_{k \ge 1}$ where $n_1 < n_2 < n_3 < \cdots$ are positive integers. If $a_n \to L$ , then every subsequence also converges to $L$ . Moreover $a_n \to L$ if and only if both $a_{2k} \to L$ and $a_{2k-1} \to L$ .

In plain words Picking out some of the terms, in order, cannot change where a convergent sequence is headed. So if two subsequences head to different places, the whole sequence diverges.

Example Let $a_n = (-1)^n \frac{n}{n + 1}$ . The even and the odd terms are

a_{2k} = \frac{2k}{2k + 1} \to 1, \qquad a_{2k-1} = -\frac{2k - 1}{2k} \to -1.

Two subsequences with different limits mean $(a_n)$ diverges.

Sequential criterion for function limits

Definition Let $f$ be defined on an open interval containing $a$ , except possibly at $a$ . Then $\lim_{x \to a} f(x) = L$ if and only if $f(x_n) \to L$ for every sequence $(x_n)$ in that interval with $x_n \ne a$ and $x_n \to a$ . For limits at infinity there is a one way version. If $\lim_{x \to \infty} f(x) = L$ and $a_n = f(n)$ for every integer $n \ge 1$ , then $a_n \to L$ .

In plain words Function limits and sequence limits talk to each other. You can borrow tools like l’Hôpital’s rule by passing to a function of a real variable, and you can disprove a function limit by finding two sequences that disagree.

Example To find $\lim \frac{\ln n}{n}$ , look at $f(x) = \frac{\ln x}{x}$ . By l’Hôpital’s rule

\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0,

so $\frac{\ln n}{n} \to 0$ . In the other direction, the limit of $\sin \frac{1}{x}$ as $x \to 0$ does not exist. Both sequences

x_n = \frac{1}{n\pi} \qquad \text{and} \qquad y_n = \frac{1}{2n\pi + \pi/2}

tend to $0$ and never equal $0$ , but $\sin \frac{1}{x_n} = 0$ for every $n$ while $\sin \frac{1}{y_n} = 1$ for every $n$ .

The converse of the one way version fails. The sequence $\sin(n\pi) = 0$ converges, but $\sin(\pi x)$ has no limit as $x \to \infty$ .

Stolz-Cesàro theorem

Definition Let $(a_n)$ and $(b_n)$ be sequences of real numbers where $(b_n)$ is strictly increasing and $b_n \to \infty$ . If

\lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = L,

where $L$ is a real number, $\infty$ or $-\infty$ , then

\lim_{n \to \infty} \frac{a_n}{b_n} = L.

In plain words It is a discrete version of l’Hôpital’s rule. To find the limit of a quotient, look at how much the top and bottom change from one step to the next.

Example Take $a_n = 1^2 + 2^2 + \cdots + n^2$ and $b_n = n^3$ , which is strictly increasing to $\infty$ . Then

\frac{a_{n+1} - a_n}{b_{n+1} - b_n} = \frac{(n + 1)^2}{3n^2 + 3n + 1} \to \frac{1}{3},

and therefore

\frac{1^2 + 2^2 + \cdots + n^2}{n^3} \to \frac{1}{3}.

The values $0.385$ , $0.33835$ and $0.3338335$ at $n = 10$ , $100$ and $1000$ agree.

Check yourself

Find $\lim_{n \to \infty} \left(\sqrt{n^2 + n} - n\right)$ .

Show the answer

Multiply and divide by the conjugate $\sqrt{n^2 + n} + n$ to get

\sqrt{n^2 + n} - n = \frac{n}{\sqrt{n^2 + n} + n} = \frac{1}{\sqrt{1 + 1/n} + 1}.

By the limit laws and continuity of the square root this tends to $\frac{1}{2}$ .

Show that $\frac{\cos(n^2)}{\sqrt{n}} \to 0$ .

Show the answer

Since $-1 \le \cos(n^2) \le 1$ , for $n \ge 1$ we have

-\frac{1}{\sqrt{n}} \le \frac{\cos(n^2)}{\sqrt{n}} \le \frac{1}{\sqrt{n}},

and both bounds go to $0$ . The squeeze theorem gives the limit $0$ .

Use the Stolz-Cesàro theorem to find $\lim \frac{1 + 2 + \cdots + n}{n^2}$ .

Show the answer

With $b_n = n^2$ strictly increasing to $\infty$ , the quotient of differences is

\frac{n + 1}{(n + 1)^2 - n^2} = \frac{n + 1}{2n + 1} \to \frac{1}{2}.

So the limit is $\frac{1}{2}$ , which matches $\frac{n(n + 1)}{2n^2}$ .

Sequence Theorems

Key terms

Uniqueness of limits

Limit laws

Comparison of limits

Squeeze theorem

Continuous functions of sequences

Subsequences

Sequential criterion for function limits

Stolz-Cesàro theorem

Check yourself

How this lesson connects

Builds on