Letting $S(m)$ be the digit sum of $m$, then $\lim_{n\to\infty}S(3^n)=\infty$?

For any $m\in\mathbb N$, let $S(m)$ be the digit sum of $m$ in the decimal system.

For example, $S(1234)=1+2+3+4=10, S(2^5)=S(32)=5$.

Question 1 :Is the following true? $$\lim_{n\to\infty}S(3^n)=\infty.$$

Question 2 :How about $S(m^n)$ for $m\ge 4$ except some trivial cases?

Motivation : I've got the following : $$\lim_{n\to\infty}S(2^n)=\infty.$$

Proof : The point of this proof is that there exists a non-zero number between the ${m+1}^{th}$ digit and ${4m}^{th}$ digit.

If $$2^n=A\cdot{10}^{4m}+B, B\lt {10}^m, 0\lt A,$$ then $2^n\ge {10}^{4m}\gt 2^{4m}$ leads $n\gt 4m$. Hence, the left side can be divided by $2^{4m}$. Also, $B$ must be divided by $2^{4m}$ because ${10}^{4m}=2^{4m}\cdot 5^{4m}$. However, since $$B\lt {10}^m\lt {16}^m=2^{4m},$$ $B$ can not be divided by $2^{4m}$ if $B\not=0$. If $B=0$, then the right side can be divided by $5$ but the left side cannot be divided by $5$. Hence, we now know that there is a non-zero number between the ${m+1}^{th}$ digit and ${4m}^{th}$ digit. Since $2^n$ cannot be divided by $5$, the first digit is not $0$. There exists non-zero number between the second digit and the fourth digit. Again, there exists non-zero number between $5^{th}$ digit and ${16}^{th}$ digit. By the same argument as above, if $2^n$ has more than $4^k$ digits, then $S(n)\ge {k+1}$. Hence, $$n\log {2}\ge 4^k-1\ \ \Rightarrow \ \ S(n)\ge k+1.$$ Now we know that $$\lim_{n\to\infty}S(2^n)=\infty$$ as desired. Now the proof is completed.

However, I've been facing difficulty for the $m=3$ case. I've got $\lim\sup S(3^n)=\infty$.

Proof : Suppose that $3^n$ has $m$ digits. Letting $l=\varphi({10}^m)+n$, then $$3^l-3^n=3^n(3^{\varphi({10}^m)}-1).$$ Since this can be divided by ${10}^m$, we know that the last $m$ digits of $3^l$ are equal to those of $3^n$. Hence, we get $\lim\sup S(3^n)=\infty$.

However, I can't get $\lim\inf S(3^n)$. Can anyone help?

Update : I crossposted to MO.

Solution 1:

I think the following result is common knowledge:

If $m$ is not a power of $10$, then for any positive integer $X$, there exists a power of $m$ which has a decimal expansion starting with $X$.

Proof idea: In other words, we should prove that $X \cdot 10^s \leq m^n < (X+1) \cdot 10^{s}$ for some positive integers $s$ and $n$. This inequality is equivalent to $s + \operatorname{log}_{10}(X) \leq n \log_{10}(m) < s + \log_{10}(X+1)$. Now the set of fractional parts of $n \alpha$, $n \in \mathbb{Z}_+$ is dense in $[0,1]$ when $\alpha$ is irrational. Using this fact we can choose the fractional part of $n \log_{10}(m)$ so that the above inequality holds (note that $\log_{10}(m)$ is irrational when $m$ is not a power of $10$).

Thus $\limsup S(m^n) = \infty$ when $m$ is not a power of $10$.

Solution 2:

I'm posting an answer just to inform that the question has received an answer by Vesselin Dimitrov on MO.

https://mathoverflow.net/questions/143263/letting-sm-be-the-digit-sum-of-m-then-lim-n-to-inftys3n-infty

He mentioned that an effective lower bound on $S(3^n)$ (going to infinity with $n$ ) is available through Baker's method; it is due to Stewart.

C. L. Stewart, On the representation of an integer in two different bases, J Reine Angew Math 319 (1980) 63-72, MR 81j:10012; H G Senge, E G Straus, PV-numbers and sets of multiplicity, Proceedings of the Washington State University Conference on Number Theory (1971) 55-67, MR 47 #8452.