# Solving the geometric series for q

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 $n\in\mathbb N$?

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.

• It is easier to solve $\frac{q^{n+1}-1}{q-1}=a$ – Peter Aug 25 '16 at 12:37
• Newton's method is often used for this case. For $a$ close to $1$, $q\approx a-1$ and for $a\gg1$, $q\approx\sqrt[n]a$. – Yves Daoust Aug 25 '16 at 12:42
• Sorry, I meant $q\approx1-1/a$, not $1-a$. – Yves Daoust Aug 25 '16 at 12:49
• For a numerical method, I wrote $a(q-1) = q^{n-1} - 1 = \int_1^q (n+1)x^n \; dx$ and then chose a $k$ and approximated the integral with trapezoid or Simpson's rule. I ended up with $a(q-1) \approx (n+1)q^{n+1}F(k)/3k^{n+1}$. Where $F(k)$ is not that hard to work out once $k$ is chosen. This near equation is only slightly easier to solve than the original, but I offer my thoughts. – B. Goddard Aug 25 '16 at 13:17
• You can find a series solution here: arxiv.org/abs/math/9411224 – N74 Aug 26 '16 at 11:32

• $$q=\frac{a-1}{a}+\delta=\frac{a-1}{a-\left( \frac{a-1}{a}+\delta \right)^{n}} \implies \frac{a-1}{a}+\delta=\frac{a-1}{a} \left[ 1+\frac{1}{a}\left( \frac{a-1}{a}+\delta \right)^{n}+\ldots \right]$$$$\implies \delta \approx \frac{(a-1)^{n+1}}{a^{n+2}}$$ and repeat in similar manner to obtain higher order terms. – Ng Chung Tak Aug 26 '16 at 12:38