Fun with math: Dividing one by 998001 yields a surprising result

> That's strange, about 998 being absent. What follows is definitely nonsense

I had to re-read what the parent poster said to get it. So I thought I'd explain what he means when he says:

> Also, "998" doesn't appear 'in sequence' (it obviously appears as 97[9 98]0 981).

He is saying 998 is missing, as others have observed. He is also saying that if you took the string of digits and did a string search for "998", you'd obviously find those digits together in other places (specifically during the sequence of 979 980 981).

You'd also find them at 799 800 801.

Slightly off-topic, but it bugs me that the article feels obliged to contain a picture of the number. Firstly, why the hell would you want to display plain text as a picture? Secondly, if I click through to the source, I get a slightly better picture. Did IHC take a photo of the original website with their phone camera and upload it? Mind is boggling. Encouraged by the better quality picture on geekosystem I decided to click through to THEIR source. And there's a plaintext version. Kudos to http://www.futilitycloset.com/2012/01/08/math-notes-76/; for having a modicum of common sense.

Here's yet another proof. This one uses probability theory.

Let's warm up with a probabilistic proof of

    1 + p + p^2 + ... = 1/(1-p)

in the form

    (1-p)(1 + p + p^2 + ...) = 1

This says that if you have a Bernoulli process with failure probability p and success probability 1-p, you will eventually succeed with probability 1. Indeed, (1-p) p^n is the probability that you will succeed after n failures.

To see this is true (it is certainly intuitively plausible), note that the probability of taking longer than n tries is p^n, which goes to zero as n goes to infinity.

We can apply the same kind of reasoning to

    (1-p)^2 (1 + 2p + 3p^2 + ...) = 1

As before, we have a process with failure probability p and success probability p-1. However, rather than stopping after the first success, we will continue until we have two successes. Now take a look at the terms of the equation's left-hand side:

    (1-p)^2 (n+1) p^n

This is the probability that we will have gotten 2 successes (and hence n failures) after n+2 steps. The second success must always be the final trial, so there are n+1 possible positions for the first success. That explains the factor of n+1.

With that understood, the equation simply says that we will eventually get 2 successes with probability 1. To see why, note that the probability of never getting any successes in n tries is p^n as before, which tends to zero. So we will eventually get at least one success with probability 1. But once we have that first success, getting the second success is just our first problem repeated, so we know that happens eventually with probability 1.

This proof technique generalizes easily. If we keep trying until we have 3 successes, we get a factor of C(n+2, 2) = (n+2)(n+1)/2 for the number of ways the first 2 successes can be placed before the third and final success when there are n failures and 3 successes. We show that a first success occurs eventually with probability 1 as before, which reduces the problem to getting two subsequent successes, a problem we already solved.

You can see this yields a proof by induction for this whole class of series identities, but it's not of those unenlightening proofs by induction that shows you something is true but does not tell you why.

For all squares, http://m.wolframalpha.com/input/?i=100010000%2F999700029999&... 100010000/999700029999 Tack on extra 0s and 9s to allocate more digits per square.