long term recurrence relations and how to recorrect the deficit

Question

I have the following question:

In a sales drive a building society is trying to gain new customers. In any 6 month period it estimates that it loses 1.5% of its customers to competitors and attracts 7000 new customers. It has 0.5 million customers at the start of its sales drive.

There are 2 questions:

a) What would happen to the number of customers in the long term if this situation continued. b) How many customers would the society have to attract in each 6 month period to maintain 0.5 million customers.

From the above, I can make the following sequence:

$${u_{n+1} = 0.985u_n + 7000}\text{ and }{u_0 = 500000}$$

Is there a better way of solving question a than using the ans function of a calculator and consistently press the equals sign to see what happens long term which to be honest is my current solution.

For question b, do I some how need to rearrange the equation to something like below and solve for a: ${u_{n+1} = 0.985u_n + a}$ and ${u_0 = 500000}$

Answer 1

In general, if $u_{n+1}=au_n +b$, then we can easily see that:

$$\begin{align} u_{n}&=b+ba+ba^2+\cdots +ba^{n-1} + a^nu_0\\ &=b\frac{1-a^n}{1-a} + a^n u_0\end{align}$$

Since $|a|<1$, the limit of this as $n$ approaches infinity is:

$$\frac{b}{1-a} = 7000/0.015 = 466666.\overline{6}$$

The (b) question is far simpler - how many do you lose in a month? You have to gain that many new customers.

Answer 2

Here's how to solve the recurrence relation $$ u_{n+1} = a u_n + b \qquad (1) $$ where $a \neq 1$. Denote $l = b / (1 - a)$ the solution of the characteristic equation $x = ax + b$ of $(1)$, that is $$ l = a l + b. \qquad \text{(2)} $$ Now, doing $(1) - (2)$ gives $$ u_{n+1} - l = a(u_n - l). $$ This shows that the sequence $(v_n)$ defined by $v_n = u_n - l$ is geometric with common ration $a$ and first term $v_0 = u_0 - l$, so $$ v_n = (u_0 - l) a^n, $$ and then $u_n = (u_0 - l) a^n + l$, or $$ u_n = \Bigl(u_0 - \frac{b}{1 - a}\Bigr) a^n + \frac{b}{1 - a}. $$