The Eternal Thing

  The divisibility rule of 3 is simple: For a given number, add the individual digits. If the sum is divisible by 3, then the overall number is divisible by 3. Otherwise, it's not. But why the sum of the digits? How can we prove that this is robust for any given number? If you wanna think it out yourself, take a pause. Otherwise, just continue reading below... The Proof: Consider a generalized decimal number, with the total number of digits as n+1, and the coefficients being a 0 , a 1 , a 2 , and so on up to a n . So, our number can be expressed as: a 0 10 0 + a 1 10 1 + a 2 10 2 + ... + a n 10 n . Now, the powers of 10 can be written in terms of 9s, right? 10 1 = 10 = 9 + 1 10 2 = 100 = 99 + 1 10 3 = 1000 = 999 + 1 and similarly, 10 n = 100..0 = 99..9 + 1. Then, our number becomes: a 0 + a 1 (9 + 1) + a 2 (99 + 1) + .... + a n (99..9 + 1) = [a 0 + a 1 + a 2 + ... + a n ] + [9*a...

  We know 1 + 1 + 1 + 1 + ..... reaches infinity.  So, in a similar way, 0.9 + 0.9 + 0.9 + 0.9 + ...... reaches infinity. What about 1/2 + 1/3 + 1/4 + 1/5 + ....?  Do you believe it reaches infinity?  If yes, you can ignore the rest of the post. If you want to try it yourself, you can close this page. But if you do not believe it reaches infinity, you are not alone!  Warning:  This post is about proving the sum as simply as possible. Please don't expect more ;) How do we want to prove it? By comparison. Say we have the value A, with us. Suppose we get another value B, with two conditions:- A > B. B reaches infinity. So, we can conclude A reaches infinity!! Job done. Splitting of our sum for comparing:- Take the first terms of our sum -  1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10. Yeah, stop at 1/10. Now, how many terms it has?  9. Now take another sum. 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 (yeah, 9 times). You c...