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?


**Remark** : This question has been asked previously on math.SE without receiving any complete answers, where spin proved that $\lim\_{n\to\infty} \sup S(m^n )=\infty$ when $m$ is not a power of $10$. 


<https://math.stackexchange.com/questions/501019/letting-sm-be-the-digit-sum-of-m-then-lim-n-to-inftys3n-infty>


**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 $\lim\_{n\to\infty}\inf S(3^n)$. Can anyone help?



Letting $S(m)$ be the digit sum of $m$, then $\lim_{n\to\infty}S(3^n)=\infty$?
This follows from W. M. Schmidt's Subspace theorem, which is a deep theorem in diophantine approximations generalizing Roth's to several variables. A full account of this theorem and its proof, as well as some of its striking applications, can be found in chapter 7 of *Heights in Diophantine Geometry* by Bombieri and Gubler. The following result, the finiteness of the number of non-degenerate solutions to the so-called "$S$-unit equation," is a straightforward application of Schmidt's theorem. (See Theorem 7.4.2 in [HIDG]):


*Let $S$ be a finite set of prime numbers, and fix $n \in \mathbb{N}$. Consider $\mathcal{X}$ the set of solutions to $x\_1 + \cdots + x\_n = 1$ in rational numbers $x\_i$ of the form $\pm \prod\_{p \in S} p^{a\_p}$, $a\_p \in \mathbb{Z}$, such that no proper subsum of $x\_1+\cdots+x\_n$ vanishes. Then $\mathcal{X}$ is a finite set.*


This implies your question immediately upon considering $S := \{2,3,5,7\}$. 


However, the proof of the Subspace theorem is not effective, and this only shows $S(3^n) \to +\infty$ without any lower estimate on the rate of growth. An effective lower bound on $S(3^n)$ (going to infinity with $n$) is available through Baker's method; it is due to Stewart, and the google search led me to the old MathOverflow post linked to in my comment below. I will just copy the references which Gerry Myerson supplied there:


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. 


