# Infinite Sequences

## Definitions

A sequence of real numbers is a function f (n), whose domain is the set of positive integers. The values an = f (n) taken by the function are called the terms of the sequence.

The set of values an = f (n) is denoted by {an}.

A sequence {an} has the limit L if for every ε > 0 there exists an integer N > 0 such that if nN, then |anL| ≤ ε. In this case we write:

$\lim\limits_{n \to \infty } {a_n} = L.$

The sequence {an} has the limit if for every positive number M there is an integer N > 0 such that if nN then {an} > M. In this case we write

$\lim\limits_{n \to \infty } {a_n} = \infty.$

If the limit $$\lim\limits_{n \to \infty } {a_n} = L$$ exists and L is finite, we say that the sequence converges. Otherwise the sequence diverges.

## Squeezing Theorem

Suppose that $$\lim\limits_{n \to \infty } {a_n} = \lim\limits_{n \to \infty } {b_n} = L$$ and $$\left\{ {{c_n}} \right\}$$ is a sequence such that $${a_n} \le {c_n} \le {b_n}$$ for all $$n \gt N,$$ where $$N$$ is a positive integer. Then

$\lim\limits_{n \to \infty } {c_n} = L.$

The sequence $$\left\{ {{a_n}} \right\}$$ is bounded if there is a number $$M \gt 0$$ such that $$\left| {{a_n}} \right| \le M$$ for every positive $$n.$$

Every convergent sequence is bounded. Every unbounded sequence is divergent.

The sequence $$\left\{ {{a_n}} \right\}$$ is monotone increasing if $${a_n} \le {a_{n + 1}}$$ for every $$n \ge 1.$$ Similarly, the sequence $$\left\{ {{a_n}} \right\}$$ is called monotone decreasing if $${a_n} \ge {a_{n + 1}}$$ for every $$n \ge 1.$$ The sequence $$\left\{ {{a_n}} \right\}$$ is called monotonic if it is either monotone increasing or monotone decreasing.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Write a formula for the $$n$$th term of $${a_n}$$ of the sequence and determine its limit (if it exists).

$\frac{1}{3},\frac{2}{4},\frac{3}{5},\frac{4}{6},\frac{5}{7}, \ldots$

### Example 2

Write a formula for the $$n$$th term of $${a_n}$$ of the sequence and determine its limit (if it exists).

$1, - \frac{2}{2},\frac{3}{4}, - \frac{4}{8},\frac{5}{{16}}, \ldots$

### Example 1.

Write a formula for the $$n$$th term of $${a_n}$$ of the sequence and determine its limit (if it exists).

$\frac{1}{3},\frac{2}{4},\frac{3}{5},\frac{4}{6},\frac{5}{7}, \ldots$

Solution.

Here $${a_n} = {\frac{n}{{n + 2}}}.$$ Then the limit is

$\lim\limits_{n \to \infty } \frac{n}{{n + 2}} = \lim\limits_{n \to \infty } \frac{{n + 2 - 2}}{{n + 2}} = \lim\limits_{n \to \infty } \left( {1 - \frac{2}{{n + 2}}} \right) = \lim\limits_{n \to \infty } 1 - \lim\limits_{n \to \infty } \frac{2}{{n + 2}} = 1 - 0 = 1.$

Thus, the sequence converges to $$1.$$

### Example 2.

Write a formula for the $$n$$th term of $${a_n}$$ of the sequence and determine its limit (if it exists).

$1, - \frac{2}{2},\frac{3}{4}, - \frac{4}{8},\frac{5}{{16}}, \ldots$

Solution.

We easily can see that $$n$$th term of the sequence is given by the formula $${a_n} = {\frac{{{{\left( { - 1} \right)}^{n - 1}}n}}{{{2^{n - 1}}}}}.$$ Since $$- n \le {\left( { - 1} \right)^{n - 1}}n \le n,$$ we can write:

$\frac{{ - n}}{{{2^{n - 1}}}} \le \frac{{{{\left( { - 1} \right)}^{n - 1}}n}}{{{2^{n - 1}}}} \le \frac{n}{{{2^{n - 1}}}}.$

Using L'Hopital's rule, we obtain

$\lim\limits_{x \to \infty } \left( { \pm \frac{x}{{{2^{x - 1}}}}} \right) = \pm \lim\limits_{x \to \infty } \frac{x}{{{2^{x - 1}}}} = \pm \lim\limits_{x \to \infty } \frac{1}{{{2^{x - 1}}\ln 2}} = 0.$

Hence, by the squeezing theorem, the limit of the initial sequence is

$\lim\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^{n - 1}}n}}{{{2^{n - 1}}}} = 0.$

See more problems on Page 2.