Alexander Bass
This note is one of many taken during 2024

Generalized Geometric Series

The geometric series is defined as

∑n=1∞rn \sum_{n=1}^{\infty} r^n

Where −1<r<1-1 < r < 1. The algebraic way to solve for the closed form xx of the series is as follows

x=r+r2+r3+…+rnx=r(1+r+r2+…+rn−1)x=r(1+x)x=r1−r \begin{align*} x &= r + r^2+r^3 + \ldots + r^n\\ x &= r(1 +r + r^2 + \ldots + r^{n-1})\\ x &= r(1+x)\\ x &= \frac{r}{1-r} \end{align*}

How about some other values for the exponent of rr? Let’s try the odd numbers 2n−12n-1

x=r+r3+r5+…+r2n−1x=r(1+r2+r4+…+r2n−2)x=r(1+r(r+r3+…+r2n−3))x=r(1+rx)x=r1−r2 \begin{align*} x &= r+r^3+r^5+\ldots + r^{2n-1}\\ x &= r(1+r^2+r^4+\ldots + r^{2n-2})\\ x &= r(1+r(r+r^3+\ldots + r^{2n-3}))\\ x &= r(1+rx)\\ x&= \frac{r}{1-r^2} \end{align*} ∑n=1∞r2n−1=r1−r2 \sum_{n=1}^{\infty} r^{2n-1} = \frac{r}{1-r^2}

How about for the multiples of three 3n3n

x=r3+r6+r9+…+r3nx=r3(1+r3+r6+…+r3n−3)x=r3(1+x)x=r31−r3 \begin{align*} x &= r^3+r^6+r^9+\ldots +r^{3n}\\ x &= r^3(1 + r^3+r^6+\ldots +r^{3n-3})\\ x &= r^3(1+x)\\ x &= \frac{r^3}{1-r^3} \end{align*} ∑n=1∞r3n=r31−r3 \sum_{n=1}^{\infty} r^{3n} =\frac{r^3}{1-r^3}

Generally, it can be assumed that for any natural number k>0k > 0 and for any real rr where −1<r<1-1 < r < 1 that

∑n=1∞rkn=rk1−rk \sum_{n=1}^{\infty} r^{kn} = \frac{r^k}{1-r^k}

To show that this is true, consider the following

x=rk+r2k+r3k+…+rnkx=rk(1+rk+r2k+…+rnk−k)x=rk(1+x)x=rk1−rk \begin{align*} x &= r^{k}+ r^{2k} + r^{3k} + \ldots + r^{nk}\\ x &= r^{k}(1+r^{k}+ r^{2k} + \ldots + r^{nk-k})\\ x &= r^{k}(1+x)\\ x &= \frac{r^{k}}{1-r^{k}} \end{align*}

How about for numbers which leave a remainder of two when divided by three: 3n−13n-1?

x=r2+r5+r8+…+r3n−1x=r2(1+r3+r6+…+r3n−3) \begin{align*} x &= r^{2} + r^{5} + r^{8} + \ldots + r^{3n-1} \\ x &= r^2 (1+ r^3 + r^6 +\ldots + r^{3n-3}) \end{align*}

The earlier found identity allows us to solve this

x=r2(1+∑n=1∞r3n)x=r2(1+r31−r3)x=r2+r51−r3x=r21−r3 \begin{align*} x &= r^2 \left(1+\sum_{n=1}^{\infty} r^{3n}\right)\\ x &= r^2\left(1+ \frac{r^3}{1-r^3}\right)\\ x &= r^2+ \frac{r^5}{1-r^3}\\ x &= \frac{r^2}{1-r^3} \end{align*} ∑n=1∞r3n−1=r21−r3 \sum_{n=1}^{\infty} r^{3n-1} = \frac{r^2}{1-r^3}

That got me thinking, is there a general closed form of

∑n=1∞ran−b=x \sum_{n=1}^{\infty}r^{an-b} = x

where a,b∈N a,b \in \mathbb{N}, 0≤b<a0 \leq b \lt a, and −1<r<1-1<r<1?

x=ra−b+r2a−b+r3a−b+…+ran−bx=ra(r−b+ra−b+r2a−b+…+ra(n−1)−b)x=ra(r−b+x)x=ra−b+raxx=ra−b1−ra \begin{align*} x&= r^{a-b} + r^{2a-b} + r^{3a-b} + \ldots + r^{an-b}\\ x&= r^{a}(r^{-b} + r^{a-b} + r^{2a-b} + \ldots + r^{a(n-1)-b}) \\ x&= r^{a}(r^{-b} +x)\\ x&= r^{a-b} +r^{a}x\\ x&= \frac{r^{a-b}}{1-r^{a}} \end{align*}

Which seems to agree with the previous result of a=3,  b=1a=3,\,\, b=1

r21−r3 \frac{r^{2}}{1-r^{3}}

And the even earlier result of odd numbers a=2,  b=1a=2,\,\, b=1

r1−r2 \frac{r}{1-r^2}

Thus I think it’s fair to say

∑n=1∞ran−b=ra−b1−ra \sum_{n=1}^{\infty}r^{an-b} =\frac{r^{a-b}}{1-r^{a}}

There’s one final case that has not been covered an+ban +b where a,b∈Na,b \in \mathbb{N} and where a,b>0a,b > 0

∑n=1∞ran+b \sum_{n=1}^{\infty} r^{an+b} x=ra+b+r2a+b+r3a+b+…+ran+bx=ra+b(1+ra+r2a+…+ra(n−1))x=ra+b(1+ra1−ra)x=ra+b+r2a+b1−rax=ra+b−r2a+b+r2a+b1−ra \begin{align*} x&=r^{a+b} + r^{2a+b} + r^{3a+b} + \ldots +r^{an+b}\\ x&=r^{a+b} (1+ r^{a} + r^{2a} + \ldots +r^{a(n-1)})\\ x&=r^{a+b} \left(1+ \frac{r^{a}}{1-r^{a}}\right)\\ x&= r^{a+b} + \frac{r^{2a+b}}{1-r^{a}}\\ x&= \frac{r^{a+b} - \cancel{r^{2a+b}} + \cancel{r^{2a+b}}}{1-r^a}\\ \end{align*}

Thus

∑n=1∞ran+b=ra+b1−ra \sum_{n=1}^{\infty} r^{an+b} = \frac{r^{a+b}}{1-r^a}

All this would seem to imply that if f(n)f(n) is a linear function which maps N→N\mathbb{N} \to \mathbb{N} then there exists a trivially obtainable closed form for the sum

∑n=1∞rf(n) \sum_{n=1}^{\infty} r^{f(n)}

I’d like to see this extended to other number systems than natural numbers, but I’m a bit exhausted of series at this point.

This note is one of many taken during 2024