Subtract from 3, Divide by 3, Repeat

There are multiple ways to approach this problem. One way is to use a calculator and finding the first few terms. The first few terms of this sequence are:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18

0
1
0.666666667
0.777777778
0.740740741
0.75308642
0.748971193
0.750342936
0.749885688
0.750038104
0.749987299
0.750004234
0.749998589
0.75000047
0.749999843
0.750000052
0.749999983
0.750000006

We see that the sequence quickly becomes very close to 0.75. Hence we can say that the limit of the sequence is $\boxed{0.75}$

Using a calculator is very easy, but the method is not very rigorous and we are not able to tell whether the limit is exactly $\frac{3}{4}$ . We will now calculate the limit using a more formal approach.

Let $A = \displaystyle \lim _{n \to \infty} a_{n }$ . We need to find the value of $A$ . Note that $\displaystyle \lim _{n \to \infty} a_{n } = \displaystyle \lim _{n \to \infty} a_{n - 1}$ . Therefore $A = \displaystyle \lim _{n \to \infty} a_{n } = \displaystyle \lim _{n \to \infty} a_{n - 1}$ . We will substitute these values in the given expression in the problem.

$A = \frac{ 3 - A } {3 }$

Solving this gives $A = \frac{ 3 } {4 } = \boxed{0.75}$ .

Another method to find the limit is to first find the explicit form of the sequence, and then finding out the limit.

We can rewrite the recurrence relation like this:

$\begin{aligned} a_{n} & = - \frac{1}{3} a_{n-1} + 1 \\ & = \left(- \frac{1}{3} \right)^{2} a_{n-2} + 1 - \frac{1}{3}\\ & = \left(- \frac{1}{3} \right)^{3} a_{n-3} + 1 - \frac{1}{3} + \frac{1}{9}\\ & = \vdots \\ & = \left(- \frac{1}{3} \right)^{n-1} a_{1} + 1 - \frac{1}{3} + \frac{1}{9} + \cdots + \left( -\frac{1}{3} \right)^{n-2}\\ \end{aligned}$

We are given $a_{1} = 0$ , so we will substitute that in the equation. We can also use the formula for sum of geometric series. $\begin{aligned} a_{n} & = 0 + \frac{1 - (-\frac{1}{3})^{n-1} }{1 - (- \frac{1}{3})} \\ & = \frac{1 - (-\frac{1}{3})^{n-1} }{\frac{4}{3}} \\ & = \frac{3}{4} \left(1 - \left(-\frac{1}{3} \right)^{n-1} \right) \\ & = \frac{3}{4} \left(1 - \left(-\frac{1}{3} \right)^{n-1} \right) \\ & = \frac{3}{4} - \frac{3}{4} \times \left(-\frac{1}{3} \right)^{n-1} \\ & = \frac{3}{4} + \frac{9}{4} \times \left(-\frac{1}{3} \right)^{n} \\ & = \frac{3}{4} + \frac{9}{4} \times (-1)^{n} \times (3) ^{-n} \\ \end{aligned}$

Now that we have obtained the explicit formula for $a_{n}$ , we can apply the limit. As $n \to \infty$ , the term $(-1)^{n}$ oscillates between 1 and -1 whereas $3^{-n}$ becomes very small approaches zero. Therefore the product approaches 0. Hence,

$\lim_{n \to \infty } a_{n} = \boxed{\frac{3}{4}} ~~~ _\square$

Let us rewrite this recurrence relation in a more suitable way.

From the original equation, we have: $a_{n} = 1 - \frac{a_{n-1}}{3} \rightarrow a_{n} + \frac{a_{n-1}}{3} = 1$ , which is true for all $n \geq 2$ . Therefore, we can write: $a_{n} + \frac{a_{n-1}}{3} = a_{n-1} + \frac{a_{n-2}}{3}$ , which in turn leads to $a_{n} - \frac{2*a_{n-1}}{3} - \frac{a_{n-2}}{3} = 0$ .

The characteristic polynomial of this recurrence relation is $q^{2} - \frac{2q}{3} - \frac{1}{3} = 0$ , that when solved yields the roots $1$ and $-\frac{1}{3}$ . Therefore, the general solution of the recurrence relation is $a_{n} = K_{1}*1^{n} + K_{2}*(-\frac{1}{3})^{n} = K_{1} + K_{2}*(-\frac{1}{3})^{n}$ .

Now, we must derive the unique solution which satisfies the problem. Although this is a second-order recurrence relation, we are only given one of its terms - $a_{1} = 0$ . This is not a problem - we can compute any other term we wish by using the original formula - for example, $a_{2} = \frac{3 - a_{1}}{3} = \frac{3 - 0}{3} = \frac{3}{3} = 1$ , and now we can solve the puzzle:

$a_{1} = K_{1} - K_{2}*\frac{1}{3} = 0$

$a_{2} = K_{1} + K_{2}*\frac{1}{9} = 1$

This yields $K_{1} = \frac{3}{4}$ and $K_{2} = \frac{9}{4}$ , and therefore the general term for $a_{n}$ is:

$a_{n} = \frac{3}{4} + \frac{9}{4}*(-\frac{1}{3})^{n}$ .

As $n$ goes to infinity, $(-\frac{1}{3})^{n}$ gets smaller and smaller, as close to 0 as it can get, which means $a_{n}$ converges to $\frac{3}{4}$ .