I found this curious exercise on the book Problems and Theorems in Analysis I, by Pólya and Szegö, and I thought it was a nice problem to propose here, since its solution requires only basic combinatorial concepts.

Consider the harmonic series $\sum_{k=1}^{+\infty} \frac{1}{k},$ which is known to diverge. Show that removing all fractions whose denominator contains the digit $$9$$ in its decimal representation leads to a convergent series.

Call $$S$$ the series obtained in the way defined above. We will try to find an upper bound to $$S$$.

1. How many numbers not containing the digit $$9$$ are there between $$1$$ and $$9$$ included? Note that all of those numbers are greater than or equal to $$1$$. How many numbers not containing the digit $$9$$ are there between $$10$$ and $$99$$ included? All these numbers are greater than or equal to $$10$$.
2. Generalyze the answer to 1.: for $$m=0,1,2,\dots$$, find a formula for the number of elements not containing $$9$$’s in the list of numbers $$\{10^{m-1}, 10^{m-1}+1,\dots, 10^{m}-1\}$$. In order to do so, recall that the first digit can be one of the numbers $$\{1,2\dots,8\}$$, whereas all the other $$m-1$$ digits can be any among $$\{0,1,\dots,8\}$$. All these numbers are greater than or equal to $$10^{m-1}$$.
3. Using the result of 2. show that $S \leq \sum_{m=1}^{+\infty} 8\cdot \left(\frac{9}{10}\right)^{m-1}.$
4. Use the fact that $\sum_{k=0}^{+\infty}a^k = \frac{1}{1-a},$valid for $$|a| < 1$$, to show that $S \leq 80.$