Three given positive $a_1, b_1, c_1$, such that $a_1+b_1+c_1=1, \forall\ n,$ $$a_{n+1}=a_n^2+2b_nc_n, b_{n+1}=b_n^2+2a_nc_n, c_{n+1}=c_n^2+2a_nb_n$$ Prove $\{a_n\},\{b_n\}$ and $ \{c_n\} $ are convergent.

I have noticed that if I add three equalities together, then get $$a_{n+1}+b_{n+1}+c_{n+1}=(a_n+b_n+c_n)^2$$ This means $\forall \ n, a_n+b_n+c_n=1$. Also, I get $$a_{n+1}-b_{n+1}=(a_1-b_1)\prod_{i=1}^n(1-3c_i).$$ But I don't know how to continue proving they are convergent? Sincerely thanks for our help.

  • $\begingroup$ Where is this problem from? $\endgroup$ – Will Fisher Jul 21 '16 at 16:44
  • $\begingroup$ @WillFisher From the exercise of a friend of mine. $\endgroup$ – Lorence Jul 21 '16 at 16:47
  • $\begingroup$ See if my new answer makes sense. $\endgroup$ – Will Fisher Jul 22 '16 at 2:23

You have already found that: $\forall \ n, a_n+b_n+c_n=1$

Without loss of generality, suppose that: $a_{n}\leq b_{n}\leq c_{n}$. Then:

$a_{n+1}=a_{n}^{2}+2b_{n}c_{n}=a_{n}^{2}+b_{n}c_{n}+b_{n}c_{n}\leq a_{n}c_{n}+b_{n}c_{n}+c_{n}c_{n}=c_{n}(a_{n}+b_{n}+c_{n})=c_{n}$.

Similary, $b_{n+1}=b_{n}^{2}+2a_{n}c_{n}\leq b_{n}c_{n}+a_{n}c_{n}+c_{n}^{2}=c_{n}(a_{n}+b_{n}+c_{n})=c_{n}$

And, $c_{n+1}=c_{n}^{2}+2a_{n}b_{n}\leq c_{n}^{2}+a_{n}c_{n}+a_{n}b_{n}=c_{n}$.

So: $\max{\{a_{n+1},b_{n+1},c_{n+1}\}}\leq c_{n}$

Had we supposed that $a_{n}$ and $c_{n}$ were bounded above by $b_{n}$ or $b_{n}$ and $c_{n}$ were bounded above by $a_{n}$ we would have gotten respectively:

$\max{\{a_{n+1},b_{n+1},c_{n+1}\}}\leq b_{n}$ or $\max{\{a_{n+1},b_{n+1},c_{n+1}\}}\leq a_{n}$

So, finally:

$\max{\{a_{n+1},b_{n+1},c_{n+1}\}}\leq \max{\{a_{n},b_{n},c_{n}\}}$

A similar argument can be used for the smallest term: supposing again that: $a_{n}\leq b_{n}\leq c_{n}$, we have:

$a_{n+1}=a_{n}^{2}+2b_{n}c_{n}\geq a_{n}^{2}+b_{n}a_{n}+c_{n}a_{n}= a_{n}$ and so on.

So we conculde that the largest $n$-th term forms a monotonic decreasing sequence which is bounded below while the smallest $n$-th term forms an increasing sequence bounded above.

This answer is in the link below where they go further to find the limit of each sequence.

Source: This problem is from Putnam Competition 1947 https://mks.mff.cuni.cz/kalva/putnam/psoln/psol475.html


This is not a complete answer but rather a hint for a attempt.

Note that $$\forall n,\;a_{n+1}+b_{n+1} = 1-c_{n+1}$$ so \begin{eqnarray*} \forall n,\;a_{n+1} &= \frac{1}{2}[(a_{n+1}-b_{n+1}) + (a_{n+1}+b_{n+1})]\\ &= \frac{1}{2}\left[(a_1-b_1)\prod_{i=1}^n(1-3c_i) + 1-c_{n+1}\right] \end{eqnarray*} and similarly \begin{eqnarray*} \forall n,\;b_{n+1} &= \frac{1}{2}[(a_{n+1}-b_{n+1}) - (a_{n+1}+b_{n+1})]\\ &= \frac{1}{2}\left[(a_1-b_1)\prod_{i=1}^n(1-3c_i) - 1+c_{n+1}\right] \end{eqnarray*}

Hence, you can plug these formulas into $c_{n+1} = c_n^2+2a_nb_n$ to obtain $$c_{n+1} = c_n^2+\frac{1}{2}\left[(a_1-b_1)\prod_{i=1}^{n-1}(1-3c_i) + 1-c_{n}\right]\left[(a_1-b_1)\prod_{i=1}^{n-1}(1-3c_i) - 1+c_{n}\right].$$

At least, this gives an expression of $c_{n+1}$ using only $c_1,\dots,c_n$ and the constants $a_1,b_1$. Hope it helps...


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