sum of squares of digits of number n

Question

I have to found a limit of $B/n$, where $B$ is a sum of squares of digits of number $n$. I was thinking about using Cesaro-Stolz theorem, but I'm stuck with the $B$.

Answer 1

Let $$n=\overline{a_k a_{k-1}\cdots a_0}=10^ka_k+\cdots +a_0$$when $a_k\ne 0$. Therefore$${B_n\over n}={a_0^2+\cdots +a_k^2\over 10^ka_k+\cdots +a_0}\le {81(k+1)\over 10^k}$$and we obtain$$0\le \lim_{n\to \infty}{B_n\over n}=\lim_{k\to \infty}{a_0^2+\cdots +a_k^2\over 10^ka_k+\cdots +a_0}\le \lim_{k\to \infty}{81(k+1)\over 10^k}=0$$therefore$$\lim_{n\to \infty}{B_n\over n}=0$$

Answer 2

Consider a number $n$ in base $b$ with $d$ digits, the high order digit being non-zero.

Then $b^{d-1} \le n < b^d$ (so that $d-1 \le \log_b(n) < d$) and $B_m(n)$, the sum of the $m$-th powers of the digits of $n$ in base $b$, satisfies

$\begin{array}\\ B_m(n) &\le d(b-1)^m\\ &\lt (\log_b(n)+1)(b-1)^m\\ \text{so}\\ \dfrac{B_m(n)}{n} &\lt \dfrac{(\log_b(n)+1)(b-1)^m}{n}\\ &= \dfrac{(\ln(n)+\ln(b))(b-1)^m}{n\ln(b)}\\ &\lt \dfrac{(2\sqrt{n}+\ln(b))(b-1)^m}{n\ln(b)} \qquad\text{since }\ln(n) < 2\sqrt{n}\\ &\lt \dfrac{3\sqrt{n}(b-1)^m}{n\ln(b)} \qquad\text{for } n > (\ln(b))^2\\ &\lt \dfrac{3(b-1)^m}{\sqrt{n}\ln(b)}\\ &\lt \epsilon \qquad\text{for } n > \left(\dfrac{3(b-1)^m}{\epsilon\ln(b)}\right)^2\\ &\to 0\\ \end{array} $

This requires $n > \dfrac{c(b, m)}{\epsilon^2} $ where $c(b, m)$ is an expression that depends on $b$ and $m$.

For any $a > 0$, this can be improved to $n > \dfrac{c(b, m, a)}{\epsilon^{1+a}} $ where $c(b, m)$ is a constant that depends on $b$, $m$, and $a$.

I don't know if this can be improved to $n > \dfrac{c_0(b, m)}{\epsilon} $ for some $c_0(b, m)$.