I have this summation series $1/a^i$, where a is a constant, and i starts at zero and goes to infinity. But the thing is, I want the sum to be 3, then solve for the value of a.
In other words, $1 + 1/a + 1/a^2 + 1/a^3$ … = 3. What is the value of a?
And hopefully someone can explain the rules how to solve for a. I don't think normal algebra rules alone will do the job, since this is an infinite series.
Optional Background:
In case you're wondering where I got this question… First of all, I'm designing an RPG game where your meters upgrade based on your level. I also know the classic assymptotic increase towards 2:
1 + 1/2 + 1/4 + 1/8 … = 2
I.e., $1/a^i$, i goes from zero to infinity, and the sum should equal 2. In this case, a equals 2. I forget what this series is called but I know it's a classical one, dating to zeno's arrow paradox I believe.
But then I thought, hmm, wouldn't it be interesting if I let your meters approach 3x their original value instead of 2? What would the series be then?
I actually found two ways to do it based on the classical one.
In the classical sum to 2, if we truncate the series and look at the values, we see 1 , 3/2 , 7/4 , 15/8, etc, all of which are closer and closer to 2 as it should be. The pattern here is $(2^{i+1} – 1)/2^i$. It was easy to extrapolate this to a sum to 3: 1 , 8/3 , 26/9 , 80/27, etc, all of which are closer and closer to 3 as they should be. The pattern there is $(3^{i+1} – 1)/3^i$.
However, to my eye, that pattern jumps a little too quickly towards 3. The jump from 1 to 8/3 = 2.667 is pretty big, so I rejected it.
The second way I found was to just go closer to 3 by 50% every time. So 1, 2, 2.5, 2.75, 2.875, etc. This doesn't "jump" like the last time, is still assymptotically approaching the value I want, and is an infinite series. However, it appears to use the exact same scheme as the classical sum to 2, just adding 1 to every term. So I consider it the same series as 1 + 1/2 + 1/4 … but just offset by a constant. I guess you could say I aesthetically don't accept it.
(In fact, I think you can choose any value greater than or equal to 50% by which you go closer to your goal. E.g., 75% would be 1, 2.5, 2.875, 2.96875 etc. I think if it's less than 50%, the series won't converge, but this still leaves a lot of choices for this scheme.)
Then I thought of the third way, $1 + 1/a + 1/a^2 + 1/a^3$ … = 3. But I don't know what a should equal.
Best Answer
Use the geometric series formula:
$$\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$$
Here, $x=\frac{1}{a}$:
$$\frac{1}{1-\frac{1}{a}} = 3$$
$$1-\frac{1}{a} = \frac{1}{3}$$
$$\frac{1}{a} = \frac{2}{3}$$
$$a = \frac{3}{2}$$