If n approaches infinity, will the descending numbers in n number system divided by the ascending numbers of n number system be closer to an integer?

Question

We know that $$\frac{987654321}{123456789}$$ is very close to $8$. If $n$ is the total amount of numbers in the number system ($10$ being the total amount of numbers in the decimal number system), will the descending numbers in $n$ number system divided by the ascending numbers of $n$ number system be even closer to an integer (e.g. $$\frac{fedcba987654321}{123456789abcdef}$$ in the hexadecimal number system)?

Answer 1

Let's call your rising number $r_n$, so $r_{\text{ten}}=123456789$, and the declining number $d_n$, so $d_{\text{ten}}=987654321$, and multiply each by $(n-1)^2$, so by $81$ in the decimal example, to see you have $$r_n = \frac{n^n -n^2+n -1}{(n-1)^2}$$ and $$d_n = \frac{(n-2)n^n +1}{(n-1)^2}$$

which makes $$\frac{d_n}{r_n} = \frac{(n-2)n^n +1}{n^n -n^2+n -1} = n-2+\frac{(n-1)^3}{n^n -n^2+n -1}.$$

The $n-2$ is the integer you are looking for, while the $\frac{(n-1)^3}{n^n -n^2+n -1}$ term declines rapidly towards $0$ as $n$ increases and is about $7.29 \times 10^{-8}$ in the decimal example

Answer 2

In base $n$, the numbers you are considering can be written as $$ a_n=\frac{p_n}{q_n},\qquad p_n=\sum_{i=1}^{n-1}in^{i-1},\qquad q_n=\sum_{i=1}^{n-1}(n-i)n^{i-1}. $$ Extrapolating from the $n=10$ case you considered would lead one to guess that $a_n\approx n-2$. First, let's simplify the fraction. Observe that $$ p_n+q_n=\sum_{i=1}^{n-1}\bigl(i+(n-i)\bigr)n^{i-1}=\sum_{i=1}^{n-1}n^i=\frac{n^n-n}{n-1}, $$ where we have summed the geometric series in the last step.

In a similar but more elaborate manner, one can also obtain a closed form for $p_n$ itself using the variant of the geometric series formula mentioned here: $$ p_n=\sum_{i=1}^{n-1}in^{i-1}=\frac{n^{n+1}-2n^n+1}{(n-1)^2} $$

Thus, we obtain the closed form $$ 1+\frac{1}{a_n}=\frac{p_n+q_n}{p_n}=(n-1)\frac{n^n-n}{(n-2)n^{n}+1}=\frac{n-1}{n-2}\cdot \frac{1-n^{1-n}}{1+n^{-n}/(n-2)}. $$ As $n\to\infty$ the right side is extremely close to $(n-1)/(n-2)$, and therefore $a_n$ grows closer and closer to $n-2$. With a little more work, from the explicit formula for $a_n$ which I have presented it can be shown that $a_n-n-2$ decays to zero more than exponentially quickly in $n$ (it will decay at an order of magnitude like $n^{-n}\approx e^{-n\log n}$).

Answer 3

We have that in a base $b$ number system, the numerator is

$$1\cdot b^0 +2\cdot b^1+3\cdot b^2+\cdots +(b-1)b^{b-2}$$

while the denominator is

$$(b-1)\cdot b^0 +(b-2)\cdot b^1+(b-3)\cdot b^2+\cdots +b^{b-2}$$

Thus, you are asking if

$$\frac{\sum_{i=1}^{b-1}i b^{i-1}}{\sum_{i=1}^{b-1}(b-i) b^{i-1}}$$

gets closer to an integer as $b$ goes to infinity? It does, and in fact converges to $b-2$. First, break apart the numerator into two parts:

$$\sum_{i=1}^{b-1}i b^{i-1}=(b-1)b^{b-2}+(b-2)b^{b-3}+\sum_{i=1}^{b-3}i b^{i-1}$$

Now, this sum has a closed form:

$$\sum_{i=1}^{b-3}i b^{i-1}=\frac{((b-4) b+2) b^b+b^3}{(b-1)^2 b^3}=\frac{-4 b^{b+1}+b^{b+2}+2 b^b+b^3}{b^5-2 b^4+b^3}=O(b^{b-3})$$

This notation simply means that as $b$ gets large, the expression behaves like $b^{b-3}$ since the largest power on the top is $b^{b+2}$ while the largest power on the bottom is $b^5$. Thus, there exists a constant $C\in\mathbb{R^{+}}$ such that

$$\left|\sum_{i=1}^{b-3}i b^{i-1}\right|\leq C b^{b-3}$$

Now, we can bound the denominator by

$$\sum_{i=1}^{b-1}(b-i) b^{i-1}\geq b^{b-2}$$

Thus

$$\lim_{b\to\infty} \left|\frac{\sum_{i=1}^{b-3}i b^{i-1}}{\sum_{i=1}^{b-1}(b-i) b^{i-1}}\right|\leq \lim_{b\to\infty} \left|\frac{Cb^{b-3}}{b^{b-2}}\right|=\lim_{b\to\infty} \frac{C}{b}=0$$

Thus, we only have to care about

$$(b-1)b^{b-2}+(b-2)b^{b-3}$$

in the numerator. In fact, we can bound the denominator by $b^{b-2}+2b^{b-3}$. If we can show

$$\lim_{b\to\infty} \left|\frac{(b-1)b^{b-2}+(b-2)b^{b-3}}{\sum_{i=1}^{b-1}(b-i) b^{i-1}}-b+2\right|\leq\lim_{b\to\infty}\left|\frac{(b-1)b^{b-2}+(b-2)b^{b-3}}{b^{b-2}+2b^{b-3}}-b+2\right|=0$$

we are done. However, this simplifies incredibly well to

$$\lim_{b\to\infty}\left|\frac{(b-1)b^{b-2}+(b-2)b^{b-3}}{b^{b-2}+2b^{b-3}}-b+2\right|=\lim_{b\to\infty}\left|\frac{2}{2+b}\right|=0$$

and the conjecture is proved.