If $a_{n+2}=(a_{n+1}+a_n)/2$ for every $n\geq1$, show that $a_n \to (a_1+2a_2)/3$.

Now, starting from the fact that $a_{n+2}-a_{n+1}=(a_n-a_{n+1})/2$ and using $b_n=a_{n+1}-a_n$ I obtain $$b_{n+1}=\frac{1}{2}b_n ⇒ b_{n+1}=\frac{1}{2^n}b_1$$ Now I sum the two sides in this way $$\sum_{k=2}^{n+1}b_k=\frac{1}{2}\sum_{k=1}^n b_k$$ Substituting back $b_k$and trying to find the limit of this sequence as $n\to\infty$ I get $$l=2a_2-a_1$$

It seems I am clearly wrong. I can prove that the sequence is Cauchy and converges (because we are in Euclidean space) but I cannot find the correct limit. Why?

Edit: Since I discovered why I got the wrong limit (read the comments), I would like to know if I could have solved this problem without using that "summation trick", which I deem not very intuitive. This sequence is an oscillating one, hence we cannot use the monotone convergence theorem.

Thank you for your help.

  • 3
    $\begingroup$ By your definition of $b_n$, you should have $b_{n+1}=-\frac{1}{2} b_n$. $\endgroup$
    – angryavian
    Commented Aug 26, 2014 at 14:01

1 Answer 1


It seems that $b_k$ has made a trouble fro you. If you use only following fact,


Then, we have $\sum_{k=1}^{n}a_{k+2}-a_{k+1}=\sum_{k=1}^n (a_k-a_{k+1})/2$, and we can conclude that $a_{n+2}-a_2=-\frac{1}{2}(a_{n+1}-a_1)$. Now, you can see that $a_n \to (a_1+2a_2)/3$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .