The $\mathbb{K}_n$ series ( or Kempner series) are obtained by removing all
terms containing a single digit $d$ from the harmonic series. For example:
$$\mathbb{K}_1= \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}+ \frac{1}{8}+ \frac{1}{9}+ \frac{1}{20}+ \frac{1}{22}+ \frac{1}{23}+\cdots$$
All $\mathbb{K}$ ten series converge as strangly as it sounds. In this topic we shall prove the convergence of $\mathbb{K}_1$.
Solution
In total there are $8 \cdot 9^k$ numbers with $k+1$ digits that do not contain $1$. (because we have $8$ choices for the first digit and $9$ for the others). Each $k+1$ digit number is less or equal to $10^k$ thus the initial series is less or equal to $\sum \limits_{k=0}^{\infty} \frac{8 \cdot 9^k}{10^k}=80$. Thus the series converges.
A similar arguement reveals the convergence of the other nine $\mathbb{K}$ series.
$$\mathbb{K}_1= \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}+ \frac{1}{8}+ \frac{1}{9}+ \frac{1}{20}+ \frac{1}{22}+ \frac{1}{23}+\cdots$$
All $\mathbb{K}$ ten series converge as strangly as it sounds. In this topic we shall prove the convergence of $\mathbb{K}_1$.
Solution
In total there are $8 \cdot 9^k$ numbers with $k+1$ digits that do not contain $1$. (because we have $8$ choices for the first digit and $9$ for the others). Each $k+1$ digit number is less or equal to $10^k$ thus the initial series is less or equal to $\sum \limits_{k=0}^{\infty} \frac{8 \cdot 9^k}{10^k}=80$. Thus the series converges.
A similar arguement reveals the convergence of the other nine $\mathbb{K}$ series.
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