1. ## a series problem

Consider the sum $\frac{1}{1}+\frac{1}{2}+...+\frac{1}{8}+\frac{1}{1 0}+...+\frac{1}{18}+\frac{1}{20}+...+\frac{1}{88}+ \frac{1}{100}+...$ where we sum over all $\frac{1}{n}$ where $n$ doesn't have the digit 9 in its decimal expansion. Show that the series converges. Hint: How many terms are there in the sum where $n$ has exactly $k$ digits?

My Work:
I see that the hint is 9^k-1 since every digit can be anything from 0-8 but not all digits can be 0. I feel like an epsilon argument might work here? But how will the number of terms help me?

2. Originally Posted by DontKnoMaff
Consider the sum $\frac{1}{1}+\frac{1}{2}+...+\frac{1}{8}+\frac{1}{1 0}+...+\frac{1}{18}+\frac{1}{20}+...+\frac{1}{88}+ \frac{1}{100}+...$ where we sum over all $\frac{1}{n}$ where $n$ doesn't have the digit 9 in its decimal expansion. Show that the series converges. Hint: How many terms are there in the sum where $n$ has exactly $k$ digits?

My Work:
I see that the hint is 9^k-1 since every digit can be anything from 0-8 but not all digits can be 0. The question seems a little confusing since it talks about the sum and then asks if the series converges.. I feel like an epsilon argument might work here? But how will the number of terms help me?
What you have is a depleted harmonic series. See here: Nick's Mathematical Puzzles: Solution 72