If you want the general formula, there's another sum to know, which is that (1 + 2 + 3 + .... + n) = n(n+1)/2.
Denoting n(n+1)/2 as "T", you get the series T * (1/n + (1/n)^2 + (1/n)^3 + ...), which isn't quite the geometric series since it's lacking a leading zero power term.
Your real answer will be the full series minus that leading term that isn't there, which is T/(1-1/n) - T, which can be simplified to n(n+1)/[2(n-1)] for any n-sided die.
Im so amazed by people who understand that stuff on a theoretical level. I'm just thinkering, like bashing numbers like a brain damaged toddler until they have an ok shape. I'm totally incapable of demonstrating the accuracy of what I do, but it...seems to work
The formula I came up with was :
(x+1)/2+y((x+1)/2)
x being the number of sides
y being the suite 1/x+1/x^2+1/x^3+[...] going on to infinity
y has a finite value (0.2) since it's a convergeant suite. I'm incapable of demonstrating why this is, but it seems to converge to 0.2, and I learnt that infinite suites that are convergeant = the value they converge to
So that gave me 3.5+3.5*0.2=4.2
But then again I have no idea how to prove that, but it works. It's very valuable information for me
An easy way of visualizing a simple case of the
geometric series is to do what one of my calc professors once called "the pizza proof". Just draw a circle, divide it in half, divide one of the halves in half, divide one of the quarters in half, and so on. You'll see that the sum of these pieces approaches 1 (the whole circle), and so at least for the case n=2 it's easy to see that 1/2 + 1/4 + 1/8 + ... = 1, and that 1 + 1/2 + 1/4 + ... = 2. This doesn't shed much light on the general formula that 1 + 1/n + 1/n^2 + .... = 1/(1 - 1/n), but it at least makes it easier to remember.
For the six sided die example, I started by writing that the average roll would be:
(1/6)(1+2+3+4+5+(6+(1/6)(1+2+3+4+5+(6+1/6(1+....)))))
-- which can be rewritten as:
(1/6)(1+2+3+4+5+6) + (1/6^2)(1+2+3+4+5+6) + ...
-- which by knowing that other sum that I posted can be generalized to:
n(n+1)/2 * [1/n + 1/n^2 + 1/n^3 + ... ]
But you need to watch out, since the geometric series includes a leading term (1/n)^0, i.e. 1, hence the other step of subtracting n(n+1)/2 from the result of the series formula.
If you want to do as Reelya suggested to avoid gaps in the distribution, you could start with imaging what the infinite average roll for say a 4 sided die would look like:
(1/4)(1+2+3+(3+1/4(1+2+3+(3+1/4(1+...)))))
-- and generalize from there.
Fakedit:
For the record, I also usually edit most of my posts like 2-6 times after posting instead of proofreading, but this tendency is balanced by a burning desire to do it quickly enough to avoid the "post last edited" mark at the bottom, which I consider to be a blemishing compromise on any post's rhetorical ethos and aesthetic integrity.
In fact, I've already edited this post twice after posting.
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