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**DontKnoMaff** Consider the sum $\displaystyle \frac{1}{1}+\frac{1}{2}+...+\frac{1}{8}+\frac{1}{1 0}+...+\frac{1}{18}+\frac{1}{20}+...+\frac{1}{88}+ \frac{1}{100}+...$ where we sum over all $\displaystyle \frac{1}{n}$ where $\displaystyle n$ 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 $\displaystyle n$ has exactly $\displaystyle k$ 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?