Let ${a_n}$ be a recursive sequence defined as follows :
$a_1 =0, a_n = \frac {{a_{n-1}} + 3}{4} , n \ge 2$.
Find the formula for the $n$th term of the sequence and find its limits .
I take $a_n = a_{n-1} = l$.
Now, $l= \frac {l + 3}{4} \implies 4l-l= 3 \implies l= 1 $
I'm getting $\lim_{{a_n} \to \infty} = l$
Is it correct? Any hints/solutions ?
On
If a limit $L$ exists, it must solve $L=\frac{L+3}{4}$ so $L=1$. At this stage we still haven't proved there is a limit, but $a_n\to 1$ iff $b_n:=1-a_n\to 0$. Indeed $b_1=1$, and for $n\ge 2$ we have $b_n=\frac{b_{n-1}}{4}$, so $b_n=\frac{1}{4^{n-1}}$ for $n\ge 1$. This proves the desired limiting behaviour, and $a_n=1-\frac{1}{4^{n-1}}$.
On
Note that $4^na_n-4^{n-1}a_{n-1}=4^n-4^{n-1}$ for all $n=2,3,\ldots$. This shows that $$4^na_n-4^1a_1=\sum_{k=2}^n\,\left(4^ka_k-4^{k-1}a_{k-1}\right)=\sum_{k=2}^n\,\left(4^k-4^{k-1}\right)=4^n-4^1\,.$$
On
Making the problem more general, consider $$a_n=\alpha\,a_{n-1}+\beta\tag 1$$ where $\alpha$ nad $\beta$ are constants.
To come back to something simpler, define $a_n=b_n+c$ and replace in $(1)$ to get $$b_n=\alpha\, b(n-1)+(\beta +\alpha c-c)\tag 2$$and let the constant term to be $0$. So, assuming $\alpha \neq 1$, $$\beta +\alpha c-c=0 \implies c=\frac{\beta }{1-\alpha }$$ and what is left is $$b_n=\alpha\, b_{n-1}\implies b_n=k\, \alpha^{n-1}\implies a_n=k\, \alpha^{n-1}+\frac{\beta }{1-\alpha }$$
Now, if $\alpha <1$, you even do not need to care about the initial condition $\alpha_0$ to show that the limit is just $\frac{\beta }{1-\alpha }$.
$a_n - a_{n-1} = \dfrac{a_{n-1}-a_{n-2}}{4}\implies b_n = \dfrac{b_{n-1}}{4}, b_n = a_n - a_{n-1}\implies b_n = \dfrac{b_{n-2}}{4^2}=...=\dfrac{b_2}{4^{n-2}}= \dfrac{a_2-a_1}{4^{n-2}}= \dfrac{3}{4^{n-1}}\implies a_n = (a_n-a_{n-1})+(a_{n-1}-a_{n-2})+\cdots+(a_2-a_1)+a_1=b_n+b_{n-1}+\cdots+b_2+a_1=\dfrac{3}{4^{n-1}}+\dfrac{3}{4^{n-2}}+\cdots+\dfrac{3}{4}+0 =....$. Can you continue?