Threefold composition of geometric series

Question

For $|r|<1$, we have the geometric series $\displaystyle{\sum_{k=0}^{\infty} r^k = \frac{1}{1-r}}$. I'll call the RHS $G(r)$ and the LHS $S(r)$. It is clear that $G(G(G(r)))=r$, even if $r=1$ with the usual convention $1/\infty=0$: $$ G(G(r)) = \frac{1}{1-\frac{1}{1-r}} = 1-\frac{1}{r};\qquad G(G(G(r))) = G(1-1/r) = \frac{1}{1-(1-1/r)}=r $$The goal: Is there a meaningful "series-side" analog of $S(S(S(r)))$? I found an interpretation of one composition. Since the series is infinitely differentiable, it's easy to see it satisfies the differential equation: $$ \frac{d^{n}}{d r^n} \frac{S(r)}{n!} = \left(S(r)\right)^{n+1} $$This gives $$ S(S(r))=1+ S(r) + S(r)^2 + S(r)^3 +S(r)^4 +S(r)^5+\cdots, $$which we can write as a sum of derivatves $$ =1+ \frac{S(r)}{1} + \frac{S'(r)}{1} + \frac{S''(r)}{2} +\frac{S'''(r)}{6} +\frac{S''''(r)}{24}+\cdots, $$which when put into the Taylor series formula reduces rather nicely $$ =1+ \sum_{n=0}^{\infty}\frac{S^{(n)}(r)}{n!} =1+ S(r+1) =1+ \frac{1}{1-(r+1)} = 1-\frac{1}{r}. $$But I haven't found a pretty way to compose once more with $S$ and get back the identity, $r$; in other words, a series proof that $S(1-1/r)=r$. Any suggestions?

Answer 1

It's no different from the usual geometric series, we have $$\sum_{n=0}^\infty\Big(1-\frac1r\Big)^n = \frac{1}{1-(1-\frac1r)}=r,$$ so long as $|1-\frac 1r|<1$.

Answer 2

There is a problem with the series analogue of composition: to formally compose two series $$ F(x) = a_0+a_1x+ \ldots \qquad G(x)=b_0+b_1x+\ldots $$ i.e. to make sense of $F(G(x))$, you need to have $b_0=0$. This is because, formally, it only makes sense to evaluate $F$ at $0$, and in order to have $G(0)=0$ you need to have $b_0=0$. In your case, $G(0)=1$, making everything very difficult.

In fact, if you tried to evaluate $G(G(r))$ you get a lot of infinite sums of positive integers: $$ 1+(1+r+\ldots)+(1+r+\ldots)^2 + (1+r+\ldots)^3+\ldots$$$$ = (1+1+1+\ldots )+ (1+2+3+\ldots )r+\ldots $$