Solving the geometric series for q

Question

Is there a general way to find the $q > 0$ solving the equation from the geometric series $$1+q+q^2+q^3+\ldots + q^n = a$$ or $$\frac{1-q^{n+1}}{1-q} = a\quad\text{with } q \neq 1$$ for $a > 1$ and $n\in\mathbb N$?

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My thoughts: Since polynomials aren't solvable in general for degree 5 or higher, I guess the above equation has no explicit solution for $n\ge 5$. In this case numerical approximations can be used. For $n=5$ also this method can be used.

Answer 1

Re-arrange \begin{align*} q &= \frac{a-1}{a-q^n} \\ &= \frac{a-1}{a-\left(\frac{a-1}{a-q^n}\right)^n} \\ &= \frac{a-1} {a-\left( \frac{a-1} {a-\left( \frac{a-1}{a-\ddots} \right)^n} \right)^n} \\ &= \frac{a-1}{a} \left \{ 1+\frac{(a-1)^n}{a^{n+1}}+\ldots+ \frac{(nk+k)!}{k! (nk+1)!} \left[ \frac{(a-1)^n}{a^{n+1}} \right]^{k}+\ldots \right \} \end{align*}

Lagrange inversion formula

\begin{align*} f(q) &= aq-q^{n+1} \\ q &= f^{-1}(a-1) \\ &= \sum_{i=1}^{\infty} \frac{1}{i!} \frac{d^{i-1}}{dx^{i-1}} \left. \left(\frac{x-0}{f(x)-f(0)} \right)^{i} \right|_{x=0} [(a-1)-f(0)]^{i} \\ &= \sum_{i=1}^{\infty} \frac{1}{i!} \frac{d^{i-1}}{dx^{i-1}} \left. \left(\frac{1}{a-x^{n}} \right)^{i} \right|_{x=0} (a-1)^{i} \end{align*}