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Let $\{ a_{n}\}$ be a sequence of positive real numbers such that $a_{n}=\sqrt{a_{n-1}a_{n-2}}$ for $n≥3$, then $\{ a_{n}\}$ converges to $(a_{1}a_{2}^{2})^{\frac{1}{3}}$.

My attempt:-

I multiplied the all new terms and simplify the terms based on the recursive relation

$a_{n}.a_{n-1}...a_{2}.a_{1}=\sqrt{a_{n-1}.a_{n-2}}.\sqrt{a_{n-2}.a_{n-3}}...a_{2}.a_{1}$

Cancel the like terms,

I get $a_{n}\sqrt{a_{n-1}}=a_{2}\sqrt{a_{1}}$.

Limit , I am getting the result. How is my steps? Does it have any mistakes? How to prove the existence?. I am not able to judge whether it is monotonically decreasing/ increasing. How to prove the sequence is bounded?

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    Hint: let $,b_n = \log a_n,$ then it reduces to $,b_n=(b_{n-1} +b_{n-2})/2,$. – dxiv Aug 10 '17 at 06:44
  • For another hint, closer to your approach: the sequence is bounded because all terms are positive and $,a_n,$ is between $,a_{n-1},$ and $,a_{n-2},$ by AM-GM. It is not monotonic, but the subsequences of odd and even indices, respectively, are monotonic, and tend to the same limit. – dxiv Aug 10 '17 at 06:52
  • @dxiv No reason to invoke AM-GM to show that it is between $a_{n-1}$ and $a_{n-2}$; $a_n$ is the side length in a square with the same area as a rectangle with sides $a_{n-1}$ and $a_{n-2}$. Besides, wouldn't AM-GM just show that $a_n$ is smaller than the largest of the two? – Arthur Aug 10 '17 at 06:55
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    @dxiv You have given again a helping hand. Thank you. –  Aug 10 '17 at 06:56
  • @Arthur The geometric mean is always between the smallest and largest of the two. – dxiv Aug 10 '17 at 06:56
  • @dxiv That is true, and I have given an outline of a proof, but invoking AM-GM as you say will only give you an upper bound. Besides, you still need to show that the AM is also between the two, so it just seems unnecessary. – Arthur Aug 10 '17 at 06:56
  • @Arthur Point taken. What I meant is that all means of $n$ variables (arithmetic, geometric, harmonic, generalized etc) lie between the smallest and largest one. For just $2$ numbers $\max(a,b)\ge(a+b)/2\ge \sqrt{ab}\ge \min(a,b),$. Maybe AM-GM wasn't the best "tag" to convey that. – dxiv Aug 10 '17 at 06:59
  • Another one here: https://math.stackexchange.com/questions/477566/show-that-the-sequence-is-convergent-to-a-1a-221-3. – Martin R Aug 10 '17 at 09:27

1 Answers1

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A first observation is that $a_n$ is of the form $$ a_n=a_1^{k_n}a_2^{1-k_n}, $$ where $k_n\in[0,1]$, with $k_1=1$, $k_2=0$, and $$ a_1^{k_{n+2}}a_2^{1-k_{n+2}}=a_{n+2}=\sqrt{a_na_{n+1}}= \sqrt{a_1^{k_{n+1}}a_2^{1-k_{n+1}}a_1^{k_{n}}a_2^{1-k_{n}}}=a_1^{\frac{k_{n}+k_{n+1}}{2}}a_2^{1-\frac{k_{n}+k_{n+1}}{2}}, $$ and hence $k_n$ satisfies $$ k_1=1,\,k_2=0,\, k_{n+2}=\frac{1}{2}(k_n+k_{n+1}). $$ Next, observe that $$ k_{n+2}-k_{n+1}=-\frac{1}{2}(k_{n+1}-k_n), $$ and hence that $$ k_{n+2}-k_{n+1}=-\frac{1}{2}(k_{n+1}-k_n)=\cdots=\frac{(-1)^n}{2^n}(k_2-k_1)=-\frac{(-1)^n}{2^n}. $$ Thus $k_n$ also satisfies $$ k_n=k_{n-1}-\frac{(-1)^{n-2}}{2^{n-2}} $$ and finally, $$ k_n=k_1-\sum_{j=0}^{n-2}\frac{(-1)^j}{2^j}=1-\frac{1-\frac{(-1)^{n-1}}{2^{n-1}}}{1-\frac{-1}{2}}\to 1-\frac{1}{1+\frac{1}{2}}=\frac{1}{3}, $$ and hence $$ a_n\to a_1^{\frac{1}{3}}a_2^{\frac{2}{3}}. $$

Yiorgos S. Smyrlis
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