# An unintuitive limit

Problem: Let $$f(n)$$ = the number of zeros in the decimal representation of $$n$$.  For example, $$f(1009)=2$$.  For $$a>0$$, define

$$S(N)=\displaystyle\sum_{k=1}^{N}a^{f(k)}$$.

Evaluate

$$L=\displaystyle\lim_{N\to\infty}\frac{\log S(N)}{\log N}$$.

Solution: The key is to recognize that the sum $$S(N)$$ is best evaluated according to how many digits are in $$N$$.  As an example, suppose $$N=9$$: none of the single-digit numbers between 1 and 9 have zeros, so the sum $$S(N)$$ is equal to 9.  Within the two-digit numbers, note that there are 9 numbers with 1 zero [e.g., 10, 20,…,90] and the rest [90-9=81] with no zeros; in this case $$S(99)=9 a+81+S(9)=9 a+9^2+S(9)$$.  For three-digit numbers, there are 3 possibilities: the number can have 2, 1, or no zeros.  There are 9 such numbers with two zeros.  Numbers with 1 zero include 101 and 110; note that there are 2 different numbers resulting from two non-zero digits and one zero digit.  Given that there are $$9^2$$ combinations of digits, it is clear that there are $$2\times 9^2$$ three-digit numbers having exactly one zero digit.  Finally, there are $$900-162-9=729=9^3$$ three-digit numbers with no zeros.  Therefore,

$$S(999)=S(10^3-1)=9 a^2+2\times 9^2 a+9^3+S(10^2-1)$$.

The pattern here is clear to the [mathematically inclined] observer:

$$S(10^k-1)=9 a^{k-1}+\binom{k-1}{1}\times 9^2 a^{k-2}+\ldots+\binom{k-1}{k-2}\times 9^{k-1} a+9^{k}+S(10^{k-1}-1)$$

or

$$S(10^k-1)=9 (a+9)^{k-1}+S(10^{k-1}-1)=9 \displaystyle\sum_{j=0}^{k-1} (a+9)^{j}$$.

Summing the series, we arrive at the following:

$$S(10^k-1)=9 \frac{(a+9)^{k}-1}{a+8}$$.

In evaluating $$L$$, observe that the limit $$N\to\infty$$ is equivalent to $$k\to\infty$$. In this limit, $$\log N \approx k$$. We therefore get the final result:

$$L=\log (a+9)$$
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