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 ?
Best Answer
$a_n - a_{n-1} = \dfrac{a_{n-1}-a_{n-2}}{4}\implies b_n = \dfrac{b_{n-1}}{4}, b_n = a_n - a_{n-1}\implies b_n = \dfrac{b_{n-2}}{4^2}=...=\dfrac{b_2}{4^{n-2}}= \dfrac{a_2-a_1}{4^{n-2}}= \dfrac{3}{4^{n-1}}\implies a_n = (a_n-a_{n-1})+(a_{n-1}-a_{n-2})+\cdots+(a_2-a_1)+a_1=b_n+b_{n-1}+\cdots+b_2+a_1=\dfrac{3}{4^{n-1}}+\dfrac{3}{4^{n-2}}+\cdots+\dfrac{3}{4}+0 =....$. Can you continue?