Limits
A linear recurrence relation is defined by \({U_{n + 1}} = a{U_n} + b\) or \({U_n} = a{U_{n - 1}} + b\)
The above relation tends to a limit as \(n \to \infty\), if and only if \(- 1\textless a \textless1\)
To find the limit we use the formula:
\(Limit = \frac{b}{{(1 - a)}}\)
Notice that the limit is not dependent on the value of \({U_1}\) or \({U_0}\).
Example
Given \({U_{n+1}}= 8U_n-0.5\) and \({U_{n+1}}= 0.5\,\,{U_n}+8\) with \({U_1}= 2\) for both, you may be asked to do any of the following:
- Calculate the first three terms for each relation
- Explain why a relation reaches a limit and give its value
- Find the value of \(n\) when a sequence exceeds a particular number
Take each of these in turn:
Calculate the first three terms for each relation
\({U_{n + 1}} = 8{U_n} - 0.5\)
\({U_1} = 2\)
\({U_2} = 8(2) - 0.5 = 15.5\)
\({U_3} = 8(15.5) - 0.5 = 123.5\)
The first three terms are \(2,15.5\) and \(123.5\)
\({U_{n + 1}} = 0.5{U_n} + 8\)
\({U_1} = 2\)
\({U_2} = 0.5(2) + 8 = 9\)
\({U_3} = 0.5(9) + 8 = 12.5\)
The first three terms are \(2,9\) and \(12.5\)
Explain why a relation reaches a limit and give its value
We have to explain why only one of these sequences approaches a limit as \(n \to \infty\). Then find the value of this limit.
\({U_{n + 1}} = 8{U_n} - 0.5\)
\(a = 8\) and \(b = - 0.5\). For limit to exist \(- 1\textless a \textless1\).
\(8\) is not between \(- 1\) and \(1\), therefore a limit does not exist for this sequence as \(n \to \infty\).
\({U_{n + 1}} = 0.5{U_n} + 8\)
\(a = 0.5\) and \(b = 8\). For limit to exist \(- 1\textless a \textless1\).
\(0.5\) is between \(- 1\) and \(1\), therefore a limit does exist for this sequence as \(n \to \infty\).
\(Limit = \frac{b}{{(1 - a)}}\)
\(= \frac{8}{{(1 - 0.5)}} = \frac{8}{{0.5}}\)
\(= \frac{{80}}{5} = 16\)
\(Limit = 16\)
Find the value of n when a sequence exceeds a particular number
For the other sequence, find the first value of \(n\) when the sequence exceeds 6000. Then find the value of this term.
\({U_{n + 1}} = 8{U_n} - 0.5\)
\({U_1} = 2\)
\({U_2} = 8(2) - 0.5 = 15.5\)
\({U_3} = 8(15.5) - 0.5 = 123.5\)
\({U_4} = 8(123.5) - 0.5 = 987.5\)
\({U_5} = 8(987.5) - 0.5 = 7899.5\)
From the calculation above:
\({U_5}\textgreater6000 \Rightarrow n = 5\)
\({U_5} = 7899.5\)
