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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

aileia
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    Interesting question but unfortunately it will likely get closed since you did not show any effort you made to solve it. – coffeemath Jul 16 '23 at 19:22
  • thaks for warning, then ı will try to write my attempt then post the question again – aileia Jul 16 '23 at 19:23
  • You should write your attempt in this present version. rather than post another. – coffeemath Jul 16 '23 at 19:25
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    Maybe not helpful to answering the question, but you might find the Kempner Series interesting if you haven't already seen it, which is the sum of the reciprocals of all natural numbers not containing a 9 digit. – mediocrevegetable1 Jul 16 '23 at 19:29
  • thanks I will have a look – aileia Jul 16 '23 at 20:05
  • The series should still divergse , but I do not see yet an easy proof for that. – Peter Jul 16 '23 at 20:19
  • 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, there 9^2 element less than 1/10, and there is 9^3 elements less than 1/100, so continues. – aileia Jul 17 '23 at 13:47
  • What counts as a 0 in the number? $1/2=0.50000...$ for example, and $1/12=0.833333$, If those must not be included, then the sum is $1/1+1/3+1/6+1/7+1/9$ – TurlocTheRed Jul 17 '23 at 14:16
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    I found a 1979 paper by Robert Baillie that gives an algorithm to compute this sum (actuall the linked paper appears to be a 2023 update of the 1979 paper). It seems that there is no known explicit formula. – TonyK Jul 17 '23 at 14:21
  • thanks ı make the computation the answer 23.1 with error margin of 0.01 – aileia Jul 17 '23 at 14:48

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