$A=\frac{1}{1}+\frac{1}{10}+\frac{1}{11}+\dots+\frac{1}{19}+\frac{1}{21}+\frac{1}{31}+\dots+\frac{1}{91}+\frac{1}{100}+\frac{1}{101}+\dots$

Question

Consider the series:

$$A=\frac{1}{1}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\dots+\frac{1}{19}+\frac{1}{21}+\frac{1}{31}+\frac{1}{41}+\dots+\frac{1}{91}+\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+\dots$$

The numerator of each term is $1$, and the denominators are the natural numbers that contain the digit $1$.

If the denominators, instead, are the natural numbers omitting numbers containing the digit $9$ is called Kempner series, it converges to $22.92067661926415034816...$.

What is the value of $A$?

Does $A$ have an exact form?

Answer 1

$$A\gt\sum_{k=1}^\infty \frac1{10k+1}\gt\sum_{k=1}^\infty\frac1{11k}=\infty $$ Hence the sum determining $A$ does not converge and $A$ does not exist.