Thread: Show convergence

if a$\displaystyle _{n+2}$ = (a$\displaystyle _{n+1}$ + a$\displaystyle _n$)/2 for all n greater or equal to 1. show that
a$\displaystyle _n$ converges to (a$\displaystyle _n$ + 2a$\displaystyle _2$)/3.

Please help me. I know that we want to show that the absolute value of a$\displaystyle _n$ minus(a$\displaystyle _n$ + 2a$\displaystyle _2$)/3 is less than epsilon but I am not sure how to get there. Please help.

Ok so am i supposed to know what the first few terms are? can somebody please help me?

The question itself makes no sense. If $\displaystyle a_n$ is sequence indexed by "n" then its limit cannot involve "n". Are you sure you have copied this correctly?

Originally Posted by janae77

if $\displaystyle a_{n+2}=\frac{ a_{n+1} + a_n}{2}$ for all n greater or equal to 1. show that
$\displaystyle a_n$ converges to $\displaystyle \frac{a_1+2a_2}{3}$.

(I corrected the question)

You can notice that the sequence $\displaystyle u_n=a_{n+1}-a_{n}$ satisfies $\displaystyle u_{n+1}=-\frac{1}{2}u_n$ and $\displaystyle u_1=a_2-a_1$ hence by induction $\displaystyle u_{n+1} = \frac{(-1)^n}{2^n}(a_2-a_1)$ (this is a geometric sequence), and you can then compute $\displaystyle a_{n+1}-a_1=u_n+u_{n-1}+\cdots+u_1=\cdots$ (sum of terms from a geometric sequence) and conclude by taking the limit as $\displaystyle n\to\infty$.