Prove that the sequence $(a_n)$ is Cauchy and find the limit. Let us define a sequence $(a_n)$ as follows:
$$a_1 = 1, a_2 = 2 \text{ and } a_{n} = \frac14 a_{n-2} + \frac34 a_{n-1}$$
Prove that the sequence $(a_n)$ is Cauchy and find the limit.

I have proved that the sequence $(a_n)$  is Cauchy. But unable to find the limit. I have observed that the sequence $(a_n)$  is decreasing for $n \ge 2$.

Given that $a_1=1$ and $a_2=2$ such that $\displaystyle a_n=\frac{1}{4}a_{n-2}+\frac{3}{4}a_{n-1}$
for $n\geq3$
Now $\displaystyle a_{n}-a_{n-1}=-\frac{1}{4}\left(a_{n-1}-a_{n-2}\right)=\left(-\frac{1}{4}\right)^2\left(a_{n-2}-a_{n-3}\right)$
$\displaystyle \dots=\left(-\frac{1}{4}\right)^{n-2}(a_2-a_1)=\left(-\frac{1}{4}\right)^{n-2}$
So  $\displaystyle \sum_{n=2}^k(a_{n}-a_{n-1})=\sum_{n=2}^k\left(-\frac{1}{4}\right)^{n-2}\Rightarrow a_k=1+\sum_{n=2}^k\left(-\frac{1}{4}\right)^{n-2}$
Now take limit we will get $\displaystyle \lim_{k\to\infty}a_k=1+\sum_{n=2}^{\infty}\left(-\frac{1}{4}\right)^{n-2}=1+\frac{4}{5}=\frac{9}{5}$
