0

Im given the sequence $a_{n+2}={a_{n}+a_{n+1}\over 2}$ where $a_{1}=0$ and $a_{2}=1$ Im asked to prove its convergent to $2\over3$.

I've looked at this for a solid 2 hours now, all I can see is that $a_{2n}>{2\over 3}$ and $a_{2n-1}<{2\over3}$ But I can't seem to prove that. Its a really weird sequence as its not monotonic. I was thinking of finding two sequences, the sequence of even n and the sequence of odd n then show they converge to 2/3 proving the sequence converges to 2/3 by squeeze.

My other thoughts were using cauchy criterion to show the terms get as close as we please.

Any thoughts appreciated.

user3258845
  • 639
  • 1
    Depends on th starting values. – Thomas Andrews Oct 10 '15 at 03:56
  • Just added them. Also the sequence will obviously be bounded between 0 and 1 – user3258845 Oct 10 '15 at 04:00
  • $a_n=\frac 2 3+ \frac 4 3(-\frac 1 2)^n$. Consequently, prove that $|a_n-2/3|$ is decreasing exponentially using recursion. – A.S. Oct 10 '15 at 04:10
  • How did you end up with that? im not sure where that came from can you explain? I was thinking since the terms were halfing each time it was related to $1/2^n$ – user3258845 Oct 10 '15 at 04:12
  • Look for solutions of the form $a_n=r^n$, solve for $r$ (two different solutions) and figure out multiplicative factors based on initial conditions. But that's more than you need. You just need to prove that $|a_n-2/3|$ is decreasing exponentially with a factor of $\frac 1 2$ and you can do that using your recurrence relation, what you already noticed (but didn't prove) and induction. – A.S. Oct 10 '15 at 04:19
  • While your sequence is not monotone, both sub-sequences (odd and even ones) are monotone. – A.S. Oct 10 '15 at 04:30

1 Answers1

1

Define for each $n\geq 1$, $b_n:=a_{n+1}-a_n$. Then $b_1=1$. Also the given recursion can be re-written as $b_{n+1}=-b_n/2$. Hence by induction for each $n\geq 0$, $b_{n+1}=(-1/2)^n b_1=(-1/2)^n$. Observe $a_{n+1}=\sum_{k=1}^n b_k=\sum_{k=1}^n (-1/2)^{k-1}$. Hence $$ \lim_{n\to\infty} a_n=\sum_{k=1}^\infty (-1/2)^{k-1} = \frac{1}{1-(-\frac{1}{2})}=2/3. $$

Raymond Reddington
  • 304