Using nth Terms of Sequences

How do I use the n^th term formula for a sequence?

nth term formulae can be given as algebraic expressions in terms of n that take positive integer values of n only
- for example, $\frac{4 n}{n + 1}$
To find the value of the first term, substitute n = 1 into the formula
To find the value of the second term, substitute n = 2 into the formula, and so on
To find which term has a value of $\frac{18}{5}$ , set the formula equal to $\frac{18}{5}$ and solve the equation to find n
- for example, $\frac{4 n}{n + 1} = \frac{18}{5}$ then solve this equation to find n = 9
  - This means the 9^th term has a value of $\frac{18}{5}$

Exam Tip

If you are asked "which term", the question usually wants to know which value of n (e.g. n = 5, so the 5th term), not its value in the sequence

Worked example

The n^th term of a sequence is given by $\frac{5 - n}{2 n^{2} + 1}$

(a)

Find the first three terms, simplifying your answers where possible.

The first three terms are when $n = 1, n = 2, n = 3$

When $n = 1$ : $\frac{5 - 1}{2 {(1)}^{2} + 1} = \frac{4}{2 + 1} = \frac{4}{3}$

When $n = 2$ : $\frac{5 - 2}{2 {(2)}^{2} + 1} = \frac{3}{8 + 1} = \frac{3}{9} = \frac{1}{3}$

When $n = 3$ : $\frac{5 - 3}{2 {(3)}^{2} + 1} = \frac{2}{18 + 1} = \frac{2}{19}$

$\frac{4}{3}, \frac{1}{3}, \frac{2}{19}$

(b)

Which term in the sequence is the first one to have a negative value?

We can see from part (a) that the terms are decreasing, and getting closer to zero (and then negative numbers)

Let's find when the sequence is equal to zero, and then after this, the sequence will be negative

$\frac{5 - n}{2 n^{2} + 1} = 0$

Multiplying both sides by the denominator and solving

$\begin{array}{rcl} 5 - n & = & 0 \\ n & = & 5 \end{array}$

So the 5^th term in the sequence is zero
As the sequence is decreasing, this means the 6^th term will be the first negative term
But we should substitute in $n = 6$ to check this

$\frac{5 - 6}{2 {(6)}^{2} + 1} = \frac{- 1}{72 + 1} = - \frac{1}{73}$

The 6^th term

Finding Limits of Sequences

What is the limiting value of a sequence?

Some sequences get closer and closer to a particular value
- This value is called the limiting value (or "limit" for short)
The sequence $\frac{n}{1 + n}$ starts 0.5, 0.66..., 0.75, 0.8, 0.83...
- the 100^th term (n = 100) is 0.990..., the 1000^th term is 0.999...
- The the limiting value of this sequence is 1
Increasing n "to infinity" finds the limiting value
- this is written " $n \to \infty$ " ("n tends to infinity")

How do I find the limiting value of a sequence?

The sequence $\frac{1}{n}$ has terms $\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, . . ., \frac{1}{1000}, . . .,$ with a limiting value of zero
- Each term gets closer and closer to zero
Similar sequences like $\frac{10}{n}$ , $\frac{1}{n^{2}}$ , $\frac{1}{n^{3}}$ , or $- \frac{4}{n^{8}}$ etc have a limiting values of zero
For n^th term formulae that are algebraic fractions in n, find the limiting value by first dividing every term (top-and-bottom) by the highest power of n
- For $\frac{6 - 2 n^{2}}{4 n^{2} + 4 n}$ divide every term by n² to get $\frac{\frac{6}{n^{2}} - 2}{4 + \frac{4}{n}}$
  - $\frac{6}{n^{2}}$ and $\frac{4}{n}$ have limiting values of zero as $n \to \infty$
  - $\frac{\frac{6}{n^{2}} - 2}{4 + \frac{4}{n}} \to \frac{0 - 2}{4 + 0}$ so the limiting value is $- \frac{2}{4}$ , i.e. $- \frac{1}{2}$
Many sequences do not have a limiting value
- The sequence $5 n$ is 5, 10, 15, 20, ... which never settles

Worked example

Find the limiting values of the following sequences given by their n^th term formulae.

(a)

\begin{matrix}  \end{matrix}

\frac{5 n + 3}{2 n - 1}

Divide the numerator and denominator by $n$

$\frac{5 + \frac{3}{n}}{2 - \frac{1}{n}}$

As $n$ tends towards infinity, $\frac{3}{n}$ tends towards 0, and $\frac{1}{n}$ tends towards 0
So the expression becomes

$\frac{5 + 0}{2 - 0}$

So the limiting value is

$\frac{5}{2}$

(b) $\begin{matrix} \end{matrix}$

$\frac{2 - n^{3}}{n^{3} + 4 n^{2} - n}$

Divide the numerator and denominator by the highest power of $n$ , which is $n^{3}$

$\frac{\frac{2}{n^{3}} - 1}{1 + \frac{4}{n} - \frac{1}{n^{2}}}$

As $n$ tends towards infinity, $\frac{2}{n^{3}}$ tends towards 0, $\frac{4}{n}$ tends towards 0, and $\frac{1}{n^{2}}$ tends towards 0
So the expression becomes

$\frac{0 - 1}{1 + 0 - 0}$

So the limiting value is

$- 1$

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

Divide the numerator and denominator by the highest power of $n$ , which is $n^{2}$

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

As $n$ tends towards infinity, $\frac{3}{n}$ tends towards 0, and $\frac{1}{n^{2}}$ tends towards 0
So the expression becomes

$\frac{0}{1 + 0}$

Notice that it is possible for the numerator to become zero
So the limiting value is

$0$

Using nth Terms of Sequences (AQA GCSE Further Maths)

Revision Note