Interesting Sum

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Yesterday I worked with what I thought was an interesting sum

Sum n = 1 --> n = K (Anr^n)

I made up my own notation here and I hope you understand it. An example would be

2*.5 + 2*2*.5^2 + 2*3*.5^3 ...

Where A is 2, and r is .5. Once you've figured out how to calculate this pattern, you can figure out what happens if you offset the linear part and the exponential part (make an n+1 somewhere for example) pretty easily.

I thought about how to do this for about an hour and found that you can do something very similar to the proof for geometric sums. I found that (S = the sum and K equals how many terms):

rS = S + A(r^k -1)/(1-r) + A + KAr^(K+1)

You can rearrange this to get

S = A[(r^K - 1)/(1-r) + Kr^(K + 1) + 1]/-(1-r)

The A(r^k -1)/(1-r) part is subtracting a regular old geometric sum. This equation is well known and I will prove it below. I put a minus sign in the denominator so that the form was 1-r so that if you want you can combine it with the /(1-r) part in the A(r^k -1)/(1-r) equation. If you set K equal to infinity then this equation becomes

S = Ar/(1-r)^2

I tested this equation by adding up about 30 terms on my calculator and I got 11.995... and the number predicted by this equation was 12, so I think I did this right. I don't know if this has been done before, but I think I'm awfully clever to figure this out by myself in only an hour.

The original proof for geometric sums goes like this:

For the sum A + Ar + Ar^2 ... + Ar^n,

Sr = Ar + Ar^2 + Ar^3... + Ar^(n+1) = S - A + Ar^(n+1)

You can rearrange Sr = S - A + Ar^(n+1) to get S = A(1-r^(n+1))/(1-r)

You will usually see the r in the numerator only taken to the n power, but this is because when in that form n is defined as the number of terms in the sum, but this n is not equal to that because you start counting at 0 instead of 1 (for the Ar^0 term)