The behaviour of the first $n$ digits of ${\underbrace{99\dots99}_{n\text{ nines}}}^{\overbrace{99\dots99}^{n\text{ nines}}}$ as $n\to\infty$

decimal-expansionelementary-number-theoryfractional-part

For a natural number $n$, let $f(n)$ denote the first $n$ digits of the decimal expansion of $${\underbrace{99\dots99}_{n\text{ nines}}}^{\overbrace{99\dots99}^{n\text{ nines}}}=(10^n-1)^{10^n-1}.$$ So we have $10^{n-1}\le f(n)<10^n$. Then what can we say about $\lim_{n\to\infty} \frac{f(n)}{10^n}$ ?

For simplicity, let $y_n=(10^n-1)^{10^n-1}$. Then we have $$y_n=10^{\log_{10}(y_n)}=\underbrace{10^{\lfloor \log_{10}(y_n)\rfloor}}_{\text{integer}}\cdot 10^{\{ \log_{10}(y_n)\}},$$

where $\lfloor\cdot\rfloor$ denotes flooring and $\{\cdot\}$ denotes the fractional part.

So the first $n$ digits of $y_n$ are given by $$f(n)=10^{n-1}\cdot 10^{\{ \log_{10}(y_n)\}}.$$

Using this (somewhat) closed form, I can do numerical experiments in Mathematica, which suggest that $\lim_{n\to\infty}\frac{f(n)}{10^n}=\frac1e$:

Indeed, it would be enough to understand the bevahiour of $\{ \log_{10}(y_n)\}$ as $n\to\infty$.

Best Answer

I figured it out: We can indeed study the asymptotic behaviour of $f(n)$ using Taylor series:

Observe that $$\log_{10}(y_n)=(10^n-1)\cdot\log_{10}(10^n-1).$$ When I first saw this, I was tempted to write $$\log_{10}(y_n)=(10^n-1)\cdot\log_{10}(10^n-1)\approx(10^n-1)\cdot n.$$

Although this approximation is very tight, I am still making an error big enough such that the first $n$ digits of $10^{\text{approximation}}$ are completely different from those of $y_n$. (Indeed, the approximation just yields an integer power of $10$.)

However, there is an easy way to quantify the error in this approximation:

\begin{split}\log_{10}(y_n)&=(10^n-1)\cdot\left(\color{orange}{\log_{10}(10^n)}+\log_{10}(10^n-1)-\color{orange}{\log_{10}(10^n)}\right)\\ &=(10^n-1)\cdot\left(n+\log_{10}\left(1-\frac1{10^n}\right)\right)\\ &=(10^n-1)\cdot\left(n-\frac{10^{-n}}{\ln(10)}+O(10^{-2n})\right)\\ &=10^n n-\frac1{\ln(10)}-n+O(10^{-n}), \end{split}

where I am using the big $O$ notation as $n\to\infty$. (The second last equality is a series of the logarithm.)

It follows that $$y_n=10^{\log_{10}(y_n)}=10^{10^n n-n}\cdot 10^{-\frac1{\ln(10)}}\cdot10^{O(10^{-n})}.$$

Notice that $$10^{-\frac1{\ln(10)}}=\exp\left(-\frac{\ln(10)}{\ln(10)}\right)=\frac1e,$$ and that $10^{O(10^{-n})}$ converges to $1$ very quickly (in fact, the speed of convergence doesn't matter, $10^{O(10^{-n})}$ could be replaced by any function that converges to $1$.) So for $n$ large enough, we have $\frac1e 10^{O(10^{-n})}\in]0,1[$.

Since $10^{10^n n-n}$ is an integer power of $10$, it follows that the first $n$ digits of $y_n$ are $$\bbox[15px,border:1px groove navy]{f(n)=\left\lfloor \frac{10^n}{e}\cdot10^{O(10^{-n})}\right\rfloor.}$$

A corollary of this result is that $$\lim_{n\to\infty} \frac{f(n)}{10^n}=\frac1e.$$

Remark 1. We don't have to consider the first $n$ digits only. For example, we also have $$\lim_{n\to\infty} \frac{y_n}{10^{10^n-n}}=\frac1e$$ and I think that we could verbatim look at for example the first $2n$ digits of $y_n$. (Or, more generally, at the first $d(n)$ digits where $d(n)$ is a sequence of natural numbers that converges to $\infty$.)

Remark 2. My procedure indicates a rather rapid convergence: Indeed we have $10^{O(10^{-n})}=\exp(O(10^{-n})\ln(10))=1+O(10^{-n})$ so that $10^{O(10^{-n})}-1$ has an asymptotic error that behaves like $10^{-n}$.

Here is a plot of the ratio of $y_n$ and the approximation given above (i.e. this is a plot of the $10^{O(10^{-n})}$ term):

Related Solutions

Value of $\lim_{n\to \infty} \frac{1}{2^n} \sum_{k=1}^{2^n} \{\log_{2}k\}$

By periodicity, $\log_2 (k)$ and $\log_2 (k) - n$ has the same fractional part. Hence \begin{align*} \lim_n \frac 1{2^n} \sum_1^{2^n} \{\log_2(k)\} &= \lim_n \frac 1{2^n} \sum_1^{2^n} \left\{\log_2 \left(\frac k{2^n}\right) \right\} \\&= \lim_n \frac 1{2^n} \sum_1^{2^n} \log_2 \left( \frac k{2^n }\right) - \left\lfloor \log_2\left(\frac k{2^n}\right)\right\rfloor \\ &= \int_0^1 \log_2(x) \mathrm dx - \lim f(n)/2^n, \end{align*} where $$ f(n) = \sum_1^{2^n} \lfloor \log_2(k/2^n)\rfloor = \sum_1^{2^n} \lfloor \log_2(k)\rfloor - 2^n \cdot n. $$ Now calculate the sum of the integer parts. Since $\lfloor \log_2(k)\rfloor = j$ when $2^j \leq k \leq 2^{j+1}-1$, we have $$ f(n) + 2^n \cdot n = 0 + n + \sum_0^{n-1} 2^j j. $$ Let $S(n) = \sum_0^{n-1}j 2^j$, then $2S(n) = \sum_0^{n-1} 2^{j+1} j = \sum_1^n 2^j (j-1)$, thus \begin{align*} S(n) = 2S(n) - S(n) &= \sum_1^n 2^j (j-1) - \sum_0^{n-1} 2^j j= \sum_2^n 2^j (j-1) - \sum_1^{n-1} 2^j j \\ &= 2^n (n-1) + \sum_2^{n-1} 2^j (-1) -2 = 2^n (n-1) -2(2^{n-1}-1) \\ &= 2^n(n-2) +2, \end{align*} then $$ \frac {f(n)}{2^n} = \frac {2^n (n-2) + 2 - 2^n n-n}{2^n} = -2 +\frac {2-n}{2^n} \to -2. $$ Hence the result is $$ \int_0^1 \log_2 (x)\mathrm d x + 2 = 2 - \frac 1 {\log(2)}. $$

[Math] Sum of digits of $a^b$ equals $ab$

I like to crack numbers but this time I'll pass. It's highly unlikely that there is any other solution. I have created a few plots of function:

$$f(a)=s(a^b)-ab$$

...for $a=2,3,4,5,6,7,8$ and $b\in [1,4000]$. All these plots look the same. The function slides towards negative infinity in a pretty linear fashion, with very little variation from the straight line. If you accept bets, I can bet a house that there are no other solutions except small ones that you already listed.

Mathematica code:

For[a=2,a<=8,a++,DiscretePlot[Total[IntegerDigits[a^b,10]]-a*b,{b,2,4000},Filling->f,Joined->False] // Print]

$a=2$

$a=3$

$a=4$

$a=5$

$a=6$

$a=7$

$a=8$

Best Answer

Related Solutions

Value of $\lim_{n\to \infty} \frac{1}{2^n} \sum_{k=1}^{2^n} \{\log_{2}k\}$

[Math] Sum of digits of $a^b$ equals $ab$

Related Question