I obtained an infinite series after solving a non-linear differential equation using Frobenius method. It is possible to obtain the coefficient for an arbitrary power of the variable, but also time consuming. In short, I do not have the $n^\text{th}$ term in terms of a known formula. My queries are the following:
(a) Is there a general method to find the coefficient of the $n^{th}$ term of an infinite series from the first few (let us say 10) terms, given that the series is convergent?
(b) The series I obtained is:
$$\frac{x^2}{6} -\frac{x^4}{9} + \frac{3x^6}{80} - \frac{71 x^8}{15120} + \frac{
10361 x^{10}}{10886400} + \frac{3539 x^{12}}{29937600} + \frac{
17111641 x^{14}}{261534873600}$$
Has anyone encountered this series? Or is there a way to find the formula for its $n^{th}$ term?
Edit 1: The differential equation in question is $-9f^2 +2f'^2 -3 f f'' + a_0 =0$. The initial conditions are $f(0)=a_0, f'(0) =0$. The Frobenius solution separates out as $f(x) = a_0 \cos^3(x) + g_0(x) + \frac{1}{a_0} g_1(x) + \frac{1}{a_0^2} g_2(x) + \dots$. I could identify the first series as $a_0 \cos^3(x)$. The series in question corresponds to $g_0(x)$.
Solution: The closed form of the function given by the series was determined by using perturbation theory methods. The solution is given by:
$g_0(x) = \frac{1}{72} \sin (x) (5 \cos (2 x)+7) \tan (x)$
Essentially you are asking after a perturbation series for large $a_0$ as then with $f(x)=a_0u(x)^3$ we get $$ 9u^6-18u^4u'^2+3(6u'^2u^4+3u''u^5)=\frac1{a_0} \iff u+u''=\frac1{9a_0u^5}\\ $$ The unperturbed equation with the initial values $u(0)=1$, $u'(0)=0$ has the solution $u(x)=\cos x$, as already found out. For the first perturbation in $u(x)=\cos x+\frac1{a_0}u_0+\frac1{a_0^2}u_1+\dots$ we get the equation $$ u_0''(x)+u_0(x)=\frac1{9\cos^5x},~~ u_0(0)=u_0'(0)=0. $$ The power series on the right starts with $\frac19(1+\frac52x^2\pm\dots)$ so that for the first coefficients one gets \begin{align} 2c_2+c_0&=\frac19&\implies c_2&=\frac1{18}\\ 6c_3+c_1&=0&\implies c_3&=0\\ 12c_4+c_2&=\frac5{18}&\implies c_4&=\frac1{54} \end{align} and so on. Then compute $g_0=3\cos^2(x)\,u_0(x)$,...