Sequences Examples 1

# Sequences Examples 1

We will now look at some examples of sequences. Be sure to check the Sequences page before reading forward.

## Example 1

Consider the sequence $\left \{ \frac{n^{2} - 3n}{3} \right \}_{n=1}^{\infty}$. Determine the first 10 terms of this sequence.

We obtain the formula $a_n = \frac{n^2 - 3n}{3}$ for any $n^{\mathrm{th}}$ term in the sequence. Therefore, plugging in $n = 1, 2, 3, ..., 9, 10$, we obtain the following terms in the sequence:

(1)
\begin{align} \quad \left \{ \frac{n^{2} - 3n}{3} \right \}_{n=1}^{\infty} = \left \{ \frac{-2}{3}, \frac{-2}{3}, 0, \frac{4}{3}, \frac{10}{3}, \frac{18}{3}, \frac{28}{3}, \frac{40}{3}, \frac{54}{3}, \frac{70}{3}, ... \right \} \end{align}

## Example 2

Determine if the sequence $\left \{ \frac{n - 1}{n^2} \right \}_{n=1}^{\infty}$ is bounded above, bounded below, or neither.

We will first calculate some terms of our sequence, that is $\left \{ \frac{n - 1}{n^2} \right \}_{n=1}^{\infty} = \left \{ 0, \frac{1}{4}, \frac{2}{9}, \frac{3}{16}, \frac{4}{25}, \frac{5}{36}, ... \right \}$. We notice that as $n \to \infty$, $a_n \to 0$, so we suspect that our sequence is bounded below by 0.

We can see this must happen as $n \to \infty$, the sequence $\{ n - 1\}_{n=1}^{\infty} < \{ n^2 \}_{n=1}^{\infty}$, so the numerator is growing at a slower rate than the denominator.

## Example 3

Determine when the sequence $\left \{ \frac{n!}{n} \right \}_{n=1}^{\infty}$ is increasing.

Recall that $n! = n \cdot (n-1) \cdot (n-2) \cdot ...2 \cdot 1$, so then $\left \{ \frac{n!}{n} \right \}_{n=1}^{\infty} = \left \{ (n - 1) \cdot (n - 2) \cdot ... \cdot 2 \cdot 1 \right \}_{n=1}^{\infty} = \left \{(n-1)! \right \}_{n=1}^{\infty}$. So we must show that $a_n < a_{n+1}$ for all $n \in \mathbb{N}$ to show this sequence is increasing.

We note that $a_n = (n - 1)! = (n-1) \cdot (n-2) \cdot ... \cdot 2 \cdot 1$ while $a_{n+1} = (n + 1 - 1)! = n! = n \cdot (n-1) \cdot (n-2) \cdot ... \cdot 2 \cdot 1$. Therefore:

(2)
\begin{align} (n - 1)! ≤ n! \\ \quad (n-1) \cdot (n-2) \cdot ... \cdot 2 \cdot 1 ≤ n \cdot (n-1) \cdot (n-2) \cdot ... \cdot 2 \cdot 1 \\ 1 ≤ n \end{align}

Since $n > 1$, $a_{1} = 0! = 1$ while $a_{2} = 1! = 1$. So for $n > 1$, our sequence is increasing.