If $a_sk^s+a_{s-1}k^{s-1}+...+a_0$ is the basis representation of $n$ with respect to the basis $k$. Then, $0<n\leq k^{s+1}-1$.

Question

If $a_sk^s+a_{s-1}k^{s-1}+...+a_0$ is the basis representation of $n$ with respect to the basis $k$. Then, $$0<n\leq k^{s+1}-1$$.

My attempt:- By basis represantation, we know that $0\leq a_j<k,j=0,1,2,3,...,s.$

$a_sk^s+a_{s-1}k^{s-1}+...+a_0<k.k^s+k.k^{s-1}+...+k=k(k^s+k^{s-1}+...+1)=k.\frac{k^{s+1}-1}{k-1}.$

I am not able to reduce further. Since $\frac{k}{k-1}>1$. Please help me.

Another attempt:- $a_sk^s+a_{s-1}k^{s-1}+...+a_0<k.k^s+k.k^{s-1}+...+k<k^{s+1}+k^s+...+k+1=\frac{k^{s+2}-1}{k-1}.$

Answer 1

The main issue with both of your attempts is that your inequalities are not precise enough. Instead, with $n \ge 0$ (note $n = 0$ can also be represented with $s = 0$ and $a_0 = 0$) in base $k$, for some $s \geq 0$ we have

$$n = \sum_{i=0}^{s}a_i k^i, \; \; 0 \le a_i \le k - 1 \; \forall \; 0 \le i \le s$$

Thus, $n$ must be less than or equal to the maximum possible value with those limits on the digits, i.e., where $a_i = k - 1$ for all $0 \le i \le s$. This then gives that

$$\begin{equation}\begin{aligned} n & \le \sum_{i=0}^{s}(k-1) k^i \\ & = \sum_{i=0}^{s}(k^{i+1}-k^i) \\ & = \sum_{i=0}^{s}k^{i+1} - \sum_{i=0}^{s}k^i \\ & = \sum_{i=1}^{s+1}k^{i} - \sum_{i=0}^{s}k^i \\ & = \left(k^{s+1}+\sum_{i=1}^{s}k^{i}\right) - \left(1 + \sum_{i=1}^{s}k^i\right) \\ & = k^{s+1} - 1 \end{aligned}\end{equation}$$