Balti Tee 2020 · Ülesanne 3

The first step is to prove, using induction, the formula

$$a_n = \frac{2^{2^n}}{3^{2^n}(b_1+b_2+\cdots+b_n)}.$$

The base case $n = 0$ is trivial. Assume the formula is valid for $a_{n-1}$, that is,

$$a_{n-1} = \frac{2^{2^{n-1}}}{3^{2^{n-1}}(b_1+b_2+\cdots+b_{n-1})}.$$

If now $a_{n-1} < \sqrt{3}$, then $b_n = 0$, and so

$$a_n = a_{n-1}^2 = \frac{2^{2^n}}{3^{2^n}(b_1+b_2+\cdots+b_{n-1})} = \frac{2^{2^n}}{3^{2^n}(b_1+b_2+\cdots+b_{n-1}+b_n)},$$

whereas if $a_{n-1} \ge \sqrt{3}$, then $b_n = \frac{1}{2^n}$, and so

$$a_n = \frac{a_{n-1}^2}{3} = \frac{2^{2^n}}{3^{2^n}(b_1+b_2+\cdots+b_{n-1})+1} = \frac{2^{2^n}}{3^{2^n}(b_1+b_2+\cdots+b_{n-1}+b_n)}.$$

This completes the induction. Next, we inductively establish the inequality $a_n \ge 1$. The base case $n = 0$ is again trivial. Suppose $a_{n-1} \ge 1$. If $a_{n-1} < \sqrt{3}$, then

$$a_n = a_{n-1}^2 \ge 1^2 = 1,$$

whereas if $a_{n-1} \ge \sqrt{3}$, then

$$a_n = \frac{a_{n-1}^2}{3} \ge \frac{(\sqrt{3})^2}{3} = 1,$$

and the induction is complete. From

$$1 \le a_n = \frac{2^{2^n}}{3^{2^n}(b_1+b_2+\cdots+b_n)} = \left( \frac{2}{3^{b_1+b_2+\cdots+b_n}} \right)^{2^n},$$

we may then draw the conclusion

$$3^{b_1+b_2+\cdots+b_n} \le 2.$$