Recursive relation

Question

Sequence defined by recursive relation $a_{n+1}=\alpha a_n +2$

Prove that if $\left|\alpha\right|\lt1,$ the sequence has a limit independent of $a_1$.

I have seen this work for the case when $\alpha=1/2$, but I'm not sure how to work out the math to generalize?

Answer 1

Clearly, if $a_n$ converges then the limit satisfies the equation $x=\alpha x+2$, and thus $x=\dfrac{2}{1-\alpha}$. We hence set $$ b_n=a_n-\frac{2}{1-\alpha}. $$ Then we have $$ b_{n+1}=a_{n+1}-\frac{2}{1-\alpha}=\alpha\,a_n+2-\frac{2}{1-\alpha} =\alpha\left(a_n-\frac{2}{1-\alpha}\right)=\alpha\,b_n. $$ Hence $$ b_n=\alpha\, b_{n-1}=\alpha^2 b_{n-2}=\cdots=\alpha^{n-1} b_1= \alpha^{n-1}\left(a_1-\frac{2}{1-\alpha}\right), $$ and thus $$ a_n=b_n+\frac{2}{1-\alpha}=\alpha^{n-1}\left(a_1-\frac{2}{1-\alpha}\right)+\frac{2}{1-\alpha}, $$ and therefore $$ \lim_{n\to\infty}a_n=\frac{2}{1-\alpha}. $$

Answer 2

Since $\alpha x+2$ is a contracting map (because of $|\alpha| < 1$ , the iteration converges to the (unique) solution of the equation

x = $\alpha x+2$

for any starting value. This follows from banachs fixpoint theorem.

The limit is given by

$$\frac{2}{1-\alpha}$$

Answer 3

$$a_{n+1} = \alpha a_n + 2$$

The first few terms in this sequence are:

$$a_1,\,\alpha a_1 + 2,\, \alpha^2 a_1 + 2\alpha + 2,\, \alpha^3 a_1 + 2 \alpha^2 + 2\alpha + 2$$

This should make it clear that the $n^{\text{th}}$ term in the sequence will be (for any $n \neq 1$):

$$a_n = \alpha^{n-1} a_1 + \displaystyle\sum\limits_{k=1}^{n-1} 2\alpha^{k-1}$$

The summation part of this explicit formula is a geometric series, and so converges for $|\,\alpha\,| < 1$. The other part goes to zero for $|\,\alpha\,| < 1$. Thus, the sequence must converge.