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 $n \ge 2$.

Rewrite $a_n$ as $$a_1=1,\ a_2=2,\ a_{n+2}=\dfrac{3}{4}a_{n+1}+\dfrac{1}{4}a_n \mathrm{\ for\ } n\geqq 1.$$
We can get
\begin{align}
&a_{n+2}-a_{n+1}=-\dfrac{1}{4}(a_{n+1}-a_{n}) \cdots (A)\\
&a_{n+2}+\dfrac{1}{4}a_{n+1}=a_{n+1}+\dfrac{1}{4}a_{n} \cdots (B)
\end{align}
Letting $b_n=a_{n+1}-a_n$, we get $b_{n+1}=-\dfrac{1}{4}b_n$ from $(A)$, thus $\{b_n \}$ is a geometric progression of ratio $-\dfrac{1}{4}$. Thus $b_n=b_1 \cdot (-\frac{1}{4})^{n-1}=(-\frac{1}{4})^{n-1}$. Therefore $$a_{n+1}-a_n=\left(-\frac{1}{4}\right)^{n-1} \cdots (C)$$
Next, letting $c_n=a_{n+1}+\dfrac{1}{4}a_n$, we get $c_{n+1}=c_n$ from $(B)$. This means that all terms of $\{c_n \}$ are equal, so $c_n=c_1=\dfrac{9}{4}$. Thus $$a_{n+1}+\dfrac{1}{4}a_n=\dfrac{9}{4} \cdots (D)$$
Calculating $(D)-(C)$, we get $\dfrac{5}{4}a_n=\dfrac{9}{4}-\left(-\dfrac{1}{4}\right)^{n-1},$ i.e., $a_n=\dfrac{9}{5}-\dfrac{4}{5}\left(-\dfrac{1}{4}\right)^{n-1}$.
Letting $n\to \infty,$ we get $\displaystyle\lim_{n\to \infty} a_n =\dfrac{9}{5}.$
