# Calculus

## Infinite Sequences and Series # Infinite Sequences

• A sequence of real numbers $$n$$ is a function $$f\left( n \right),$$ whose domain is the set of positive integers. The values $${a_n} = f\left( n \right)$$ taken by the function are called the terms of the sequence.

The set of values $${a_n} = f\left( n \right)$$ is denoted by $$\left\{ {{a_n}} \right\}.$$

A sequence $$\left\{ {{a_n}} \right\}$$ has the limit $$L$$ if for every $$\varepsilon \gt 0$$ there exists an integer $$N \gt 0$$ such that if $$n \ge N,$$ then $$\left| {{a_n} – L} \right| \le \varepsilon .$$ In this case we write:

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

The sequence $$\left\{ {{a_n}} \right\}$$ has the limit $$\infty$$ if for every positive number $$M$$ there is an integer $$N \gt 0$$ such that if $$n \ge N$$ then $${a_n} \gt M.$$ 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 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 3

Determine whether the sequence $$\left\{ {\large\frac{{2n + 3}}{{5n – 7}}\normalsize} \right\}$$ converges or diverges.

### Example 4

Does the sequence $$\left\{ {\large\frac{{{n^2}}}{{{2^n}}}\normalsize} \right\}$$ converge or diverge?

### Example 5

Determine whether the sequence $$\left\{ {\sqrt {n + 2} – \sqrt {n + 1} } \right\}$$ converges or diverges.

### Example 6

Determine whether the sequence $$\left\{ {\large\frac{{5n – 7}}{{3n + 4}}\normalsize} \right\}$$ is increasing, decreasing, or neither.

### Example 7

Determine whether the sequence $$\left\{ {\large\frac{{{2^n} + 3}}{{{2^n} + 1}}\normalsize} \right\}$$ is increasing, decreasing, or not monotonic.

### 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 $${a_n} = {\large\frac{n}{{n + 2}}\normalsize}.$$ 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 $${a_n} = {\large\frac{{{{\left( { – 1} \right)}^{n – 1}}n}}{{{2^{n – 1}}}}\normalsize}.$$ 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

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

Page 1
Problems 1-2
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Problems 3-7