I am given the recursive sequence: $$ a_{n+1} = \frac{\lambda}{4} a_n +1 ,\qquad a_0 = 1. $$ I am asked to find for which $\lambda$ the sequence converges, the 4 options being:
- $\lambda > 4$,
- $\lambda < 1$,
- $\lambda > -1$,
- $|\lambda| < 4$.
I know that the limit of the sequence is $\frac {4}{4-\lambda}$ but I have no idea how to proceed from here. There are no examples with a parameter in my lesson's ppt slides, no useful theorems, and really nothing helpful on the internet at all.
The only assumption I'm somewhat certain of is that option $3$ cannot be correct, as it implies λ can be 4, which would mean the sequence diverges. However I have no idea how to interpret the other options and check them.
this is my first ans on this website so I am not so sure how to use math symbols.
This question can be solved if you find the alternative form of $a_{n+1}$. I did this by finding $a_1, a_2, a_3, a_4$.. etc.
OK so here is the answer: The sequence $a_{n+1}$ = $\sum(\lambda/4)^r$ from 0 to n+1 forming a GP. Now, we you knownthst GP converges when $\lambda/4$ < |1|.
Giving us option D.