What is the summation of reciprocals according to multiplication of natural numbers which does not have 0 as a digit? $$ S = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}..+\frac{1}{9}+\frac{1}{11}+...+\frac{1}{99}+\frac{1}{111}... $$ I try to think first all zeros were there $$\sum \frac{1}{i}$$ it is ln(X) so it goes to infinity I try to delete multiples of 10 $$\sum \frac{1}{i}-\sum \frac{1}{10*i}$$ still infinity by this type of elimination and I sandwiched it between two series. a number between 21.88 and 25.433 can be answer, and ı was expecting a solution closer to lower limit
the series do not diverge, ı can find a series bounding above that their sum is convergent. sum_(ı=0 to infty) 9(9/10)^i is an upper bound. there is 9 element less than 1 in 1/1+1/2.+1/9, there 9^2 element less than 1/10in 1/11+1/11...1/99, and there is 9^3 elements less than 1/100 in 1/111+1/112...+1/999, so continues. by this logic (sum 1/n, n=1,9)+(sum (sum 1/(kx10+n),n=1,9),k=1,9)(sum (9/10)^i,i=0,infty) is also an upper bound which add up to 23.4841 (sum 1/n, n=1,9)+(sum (sum 1/(kx10+n),n=2,10),k=1,9)(sum (9/10)^i,i=0,infty) is a lower bound which add up to 22.72
ı will try to write a more elaborate try later but, I didn't want my question to be deleted