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?

  • 1
    $\begingroup$ 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, $x \leq \dfrac{1}{3}$. $\endgroup$ Mar 2, 2022 at 18:23

1 Answer 1


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 $$\sum_{k=1}^\infty \frac{a_k}{p^k} = \frac{a_1}p + \frac 1p\sum_{k=1}^\infty \frac{a_{k+1}}{p^k}$$


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