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 $n \ge 2$.
 A: 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 $\displaystyle \lim_{k\to\infty}a_k=1+\sum_{n=2}^{\infty}\left(-\frac{1}{4}\right)^{n-2}=1+\frac{4}{5}=\frac{9}{5}$
A: For the limit I would treat it a second-order difference equation:
$$
4a_{n} -3a_{n-1} - a_{n-2}=0
$$
Conjecture the solution is $a_{n}=C_{1}b^{n}$. Then $b=1$ or $b=\frac{1}{4}$. So
$$
a_{n}=C_{1}+C_{2}\left(-\frac{1}{4}\right)^{n}
$$
Using the initial conditions
$$4 = 4C_{1}-C_{2}$$
$$32 = 16C_{1}+C_{2}$$
Solve for $C_{1}$ and $C_{2}$. Finally, notice that since $\frac{1}{4}<1$ then $\lim_{n\to\infty}a_{n}=C_{1}$. In this setting $C_{1}=\frac{9}{5}$
A: Note
$a_{n} = \frac{9}{5}+\frac{16}{5}(\frac{-1}{4})^{n}$
Thus the limit is $\frac{9}{5}$.
A: Using
$$\displaystyle a_{n}-a_{n-1}=-\frac{1}{4}\left(a_{n-1}-a_{n-2}\right)$$
one has, for $\forall n, p\in\mathbb{N}$,
$$ |a_{n+p}-a_n|=|\sum_{k=0}^{p-1}(a_{n+k+1}-a_{n+k})|\le\sum_{k=0}^{p-1}\bigg(\frac{1}{4}\bigg)^{k}|a_{n+1}-a_n|\le\bigg(\frac{1}{4}\bigg)^{n}|a_2-a_1|. $$
This implies that $\{a+n\}$ is Cauchy.
A: 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 $\displaystyle\lim_{n\to \infty} a_n =\dfrac{9}{5}.$
