# Calculus

## Infinite Sequences and Series # Infinite Sequences

## Solved Problems

Click or tap a problem to see the solution.

### Example 3

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

### Example 4

Does the sequence $\left\{ {\frac{{{n^2}}}{{{2^n}}}} \right\}$ 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 $\left\{ {\frac{{5n - 7}}{{3n + 4}}} \right\}$ is increasing, decreasing, or neither.

### Example 7

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

### Example 3.

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

Solution.

Divide by the highest power in the numerator and denominator:

$\lim\limits_{n \to \infty } \frac{{2n + 3}}{{5n - 7}} = \lim\limits_{n \to \infty } \frac{{\frac{{2n + 3}}{n}}}{{\frac{{5n - 7}}{n}}} = \lim\limits_{n \to \infty } \frac{{2 + \frac{3}{n}}}{{5 - \frac{7}{n}}} = \frac{2}{5}.$

Hence, the sequence converges to $$\frac{2}{5}.$$

### Example 4.

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

Solution.

As L'Hopital's rule yields

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

Since the limit is finite, the given sequence converges.

### Example 5.

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

Solution.

We multiply this expression by the quotient

$\frac{{\sqrt {n + 2} + \sqrt {n + 1} }}{{\sqrt {n + 2} + \sqrt {n + 1} }} = 1$

to obtain:

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

This means that the sequence converges.

### Example 6.

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

Solution.

The $$\left( {n + 1} \right)$$th term of the sequence is given by the formula

$a_{n + 1} = \frac{{5\left( {n + 1} \right) - 7}}{{3\left( {n + 1} \right) + 4}} = \frac{{5n - 2}}{{3n + 7}}.$

Check the inequality $${a_n} \le {a_{n + 1}}:$$

$\frac{{5n - 7}}{{3n + 4}} \le \frac{{5n - 2}}{{3n + 7}},\;\; \Rightarrow \frac{{5n - 7}}{{3n + 4}} - \frac{{5n - 2}}{{3n + 7}} \le 0,\;\; \Rightarrow \frac{{ - {41}}}{{\left( {3n + 4} \right)\left( {3n + 7} \right)}} \le 0.$

The last inequality is obvious, since the numerator is negative and $$3n + 4 \ge 0$$ and $$3n + 7 \ge 0$$ for $$n \ge 1.$$ Therefore, this sequence is increasing.

### Example 7.

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

Solution.

Write out the first few terms of the sequence:

$\left\{ {\frac{{{2^n} + 3}}{{{2^n} + 1}}} \right\} = \left\{ {\frac{5}{3},\frac{7}{5},\frac{{11}}{9},\frac{{19}}{{17}},\frac{{35}}{{33}}, \ldots } \right\}.$

We see that this is a decreasing sequence. To confirm this, we prove the inequality $${a_n} \ge {a_{n + 1}}.$$ We have

${a_n} = \frac{{{2^n} + 3}}{{{2^n} + 1}},\;\; {a_{n + 1}} = \frac{{{2^{n + 1}} + 3}}{{{2^{n + 1}} + 1}}.$

Then the condition $${a_n} \ge {a_{n + 1}}$$ implies that

$\frac{{{2^n} + 3}}{{{2^n} + 1}} \ge \frac{{{2^{n + 1}} + 3}}{{{2^{n + 1}} + 1}}.$

Multiply both sides of the inequality by $$\left( {{2^n} + 1} \right)$$ $$\left( {{2^{n + 1}} + 1} \right):$$

$\left( {{2^n} + 3} \right)\left( {{2^{n + 1}} + 1} \right) \ge \left( {{2^{n + 1}} + 3} \right)\left( {{2^n} + 1} \right),\;\; \Rightarrow {2^n}{2^{n + 1}} + 3 \cdot {2^{n + 1}} + {2^n} + 3 \ge {2^n}{2^{n + 1}} + 3 \cdot {2^n} + {2^{n + 1}} + 3,\;\; \Rightarrow 2 \cdot {2^{n + 1}} \ge 2 \cdot {2^n},\;\; \Rightarrow {2^{n + 1}} \ge {2^n},\;\;\Rightarrow 2 \ge 1.$

Since the last inequality is true, we can conclude that the sequence is decreasing.