sum of squares of digits of number n 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 $B$.
 A: 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 $c_0(b, m)$.
A: 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$$\lim_{n\to \infty}{B_n\over n}=0$$
