(Uniform Series Compound Amount - Annuity Converts a uniform amount (annuity) - to a future value)
Find $i$ by formula ${F \over A}={{(1+i)^n-1}\over{i}}$
where
$F$ = future value
$A$ = uniform amount per period
$i$ = interest rate
$n$ = numbers of periods
My attempt:
${F \over A}={{(1+i)^n-1}\over{i}}$
${F \over A}={{[1+ni+{{n(n-1)i^2}\over{2!}}+ {{n(n-1)(n-2)i^3}\over{3!}}+...]-1}\over{i}}$
${F \over A}={{ni+{{n(n-1)i^2}\over{2!}} +{{n(n-1)(n-2)i^3}\over{3!}}+...}\over{i}}$
${F \over A}={{i[n+{{n(n-1)i}\over{2!}} +{{n(n-1)(n-2)i^2}\over{3!}}+...]}\over{i}}$
${F\over A}= n+{{n(n-1)i}\over{2!}}+ {{n(n-1)(n-2)i^2}\over{3!}}+...$
What can I do now?
You cannot get an explicit solution for $i$ and you need some numerical method (such as Newton) to solve of the equation $$a={{(1+i)^n-1}\over{i}}$$ You can consider that you look for the zero of function $$f(x)=(1+i)^n-a\,i-1$$ $$f'(i)=n(1+i)^{n-1}-a$$ and Newton iterates would given by $$i_{k+1}=i_k-\frac{f(i_k)}{f(i_k)}$$ This should converge quite fast starting from a "reasonable" estimate $i_0$.
As you did, if you want an approximation, since $i$ is small, use the Taylor expansion built at $i=0$ (you wrote it) to get $$a=n+\frac{n(n-1) }{2} \,i+\frac{n(n-1)(n-2)}{6} i^2 +\frac{n(n-1)(n-2)(n-3)}{24} i^3 +O\left(i^4\right)$$ and use series reversion to get $$i=\frac{2 (a-n)}{(n-1) n}+\frac{4 (2-n) (a-n)^2}{3 (n-1)^2 n^2}+\frac{2 \left(5 n^2-17 n+14\right) (a-n)^3}{9 (n-1)^3 n^3}+O\left((a-n)^4\right)$$ To make it looking nicer, define $x=\frac{2(a-n)}{n(n-1)}$ and get, as an approximation, $$i=x+\frac{2-n}{3} x^2+\frac{5 n^2-17 n+14}{36} x^3+O\left(x^4\right)\tag 1$$
For illustration, let us try with $a=30$, $n=20$ and $i_0=x=\frac 1 {19}$; the iterates would be $$\left( \begin{array}{cc} n & i_n \\ 0 & 0.052631579 \\ 1 & 0.043476963 \\ 2 & 0.040921127 \\ 3 & 0.040716395 \\ 4 & 0.040715105 \end{array} \right)$$ Uisng $(1)$, we should get $i \approx \frac{587}{13718}\approx 0.0428$ which could be a better starting point for Newton method. But, since we need to iterate, the simplest is just to start using $i_0=x=\frac{2(a-n)}{n(n-1)}$.
However, we can improve quite much the estimate pushing the Taylor series to high order and to transform the resulting series to an $[n+1,n]$ Padé approximant. This would write $$i_{(n)}=x \frac{c_0+\sum_{k=1}^n c_k x^k } {c_0+\sum_{k=1}^n d_k x^k }$$ where the coefficients $c_k$ and $d_k$ are polynomials in $n$. For example, $$i_{(0)}=x$$ $$i_{(1)}=x \frac{12+(n+1) x } {12+(5 n-7) x }$$ $$i_{(2)}=x \frac{(1980 n-2340)+\left(720 n^2-1080 n+360\right) x+\left(16 n^3-12 n+4\right) x^2 } {(1980 n-2340)+\left(1380 n^2-3180 n+1920\right) x+\left(201 n^3-720 n^2+873 n-366\right) x^2 }$$ and so on.
Applied to the above example, this would give $$i_{(0)}=\frac 1 {19}\approx 0.0526316\qquad i_{(1)}=\frac{83}{2033}\approx 0.0408264\qquad i_{(2)}=\frac{345316}{8481049}\approx 0.0407162$$ to be compared to the exact solution $0.0407151$.