TITLE: Sum over all non-evil numbers
QUESTION [7 upvotes]: I'm working on the following contest math problem:

Define an evil number to be any positive integer that contains the
  digit $9$. Show that
$$ \sum_{x} \frac{1}{x} < 80 $$
where the sum is over all non-evil positive integers $x$.

I'm very confused on where to begin. Initially, I tried to consider this sum as part of the sum of $1/x$ over all positive integers, namely by noting that
$$
\sum_{n=1}^{\infty} \frac{1}{n} = \sum_{x} \frac{1}{x} + \sum_{y} \frac{1}{y}
$$
where the first summation on the right side is the same as in the problem statement and the second is the sum over all evil numbers. However, I can't seem to find a way to use this since the sum of the left diverges.
Could anyone lend a helping hand?

REPLY [3 votes]: There are $8\cdot9^{k-1}$ $k$ digit numbers without a $9$. Even if we take $8\cdot9^{k-1}$ times the reciprocal of the smallest $k$ digit number, we get the overestimate
$$
\begin{align}
\sum_{j=10^{k-1}}^{10^{k-1}+8\cdot9^{k-1}-1}\frac1j
&\le8\left(\frac9{10}\right)^{k-1}
\end{align}
$$
Add these up for $k\ge1$, we get
$$
\begin{align}
\sum_{k=1}^\infty8\left(\frac9{10}\right)^{k-1}
&=8\frac1{1-\frac9{10}}\\
&=80
\end{align}
$$