# Ternary sequence

Consider $$p\in\mathbb{N}$$, $$p\ge 2$$ we know that fixed $$x\in[0,1)$$ exists a sequence $$\{a_k\}\subseteq\mathbb{N}$$ such that for each $$k\in\mathbb{N}$$ we have $$0\le a_k\le p-1$$ and$$x=\sum_{k=1}^{\infty}\frac{a_k}{p^k}$$

We fix $$p=3$$ and $$x\in \big(\frac{1}{3},\frac{2}{3}\big)$$, then $$x=\sum_{k=1}^{\infty}\frac{a_k}{3^k},\quad a_k\in\{0,1,2\}$$ I have to show that necessarily $$a_1=1$$.

I have made several attempts even increasing with the geometric series, but I cannot conclude anything, could someone give me a suggestion?

• Hint : The case $a_1=2$ is easily excluded because $x < \dfrac{2}{3}$. To exclude the case $a_1=0$, compute $$\sum_{k=2}^{+\infty} \dfrac{2}{3^k}$$ and deduce that if $a_1=0$, then necessarily, $x \leq \dfrac{1}{3}$. Mar 2, 2022 at 18:23

First show that for any sequence $$\{b_k\}$$ with $$0 \le b_k \le p-1$$, that $$\sum_{k=1}^\infty \frac{b_k}{p^k}$$ converges to a value between $$0$$ and $$1$$. This can be done by noting that the summands are all positive, so the sequence of partial sums is increasing, and is bounded above by $$1$$.
Then notice that $$\sum_{k=1}^\infty \frac{a_k}{p^k} = \frac{a_1}p + \frac 1p\sum_{k=1}^\infty \frac{a_{k+1}}{p^k}$$