# Limit of an arithmetic average series

Sorry in advance as English is not my primary language.

I randomly thought of the following simple problem, and I coudn't solve it after one one hour trying. Maybe you guys can help.

Let $$a_1$$ and $$a_2$$ be positive real numbers. Let $$a_n$$ be the arithmetic average of the previous 2 numbers, i.e.:

$$a_n = \frac{a_{n-1}+a_{n-2}}{2}$$

If I draw this as points on a paper, I can obviously see that the limit of $$a_n$$ as $$n\rightarrow\infty$$ is a function of $$a_1$$ and $$a_2$$, but I can't solve it.

How do I proceed?

• Did you solve the recurrence equation ? If you did, did you apply the conditions ? By the way, welcome to the site ! – Claude Leibovici Feb 8 at 11:42
• You can try with, for example $a_0 = 0$ and $a_1 =100$ and plot it! It gives an interesting graphical solution ... – Matti P. Feb 8 at 11:42
• – robjohn Feb 8 at 12:05
• @ClaudeLeibovici, Thank you! I didn't even know this was a "recurrence equation", I'm not a mathematician, just an Engineer. – Anderson Linhares Feb 8 at 15:33
• @MattiP., You are right, that was helpful. I fount out that the limit as $n\rightarrow\infty$ is $\frac{a_1+2a_2}{3}$. However, I still can't find the solution for a general $a_n$. – Anderson Linhares Feb 8 at 15:33

## 1 Answer

If you rewrite the equation as

• $$2a_n -a_{n-1} - a_{n-2} = 0$$

you obtain a so called homogeneous linear difference equation.

This type of equation can be solved.

A possible method is to check what happens if you plug in a "guessed" solution of the form $$a_n = c\cdot \lambda^n$$ where $$c$$ is a real constant.

You will find that the solution can be written as $$a_n = c_1\cdot 1^n + c_2\cdot \left(-\frac{1}{2} \right)^n = c_1 + c_2\cdot \left(-\frac{1}{2} \right)^n$$

The $$1$$ and $$-\frac{1}{2}$$ come from solving the quadratic equation $$2\lambda^2 - \lambda - 1 = 0$$ you will come across when you carry out the suggested approach $$a_n = c\cdot \lambda^n$$.

• Thank you for this insight. This is not the solution though, right? Experimenting with the values, I'm certain that the limit as $n\rightarrow\infty$ is $\frac{a_1+2a_2}{3}$. I don't know what solution to "guess" to get an expression for $a_n$ though. – Anderson Linhares Feb 8 at 15:25
• it is. it iis. you may use your two starting values to determine the constants $c_1$ and $c_2$. i thought to leave this last step for you so you can play around with the general formula. – trancelocation Feb 8 at 15:29