Interval of convergence of Lagrange's infinite series I am reading a book on Orbital Mechanics for Engineering Students by Howard D. Curtis. In that book it was mentioned (in page 119) that there is no closed form solution for $E$ as a function of the eccentricity $e$ in the equation $E-e\sin E=M_e$ but there exists infinite series solutions one of which is given by
$$E = M_e+ \sum_{i=0}^{\infty} a_n e^n$$ where
$$a_n= 2^{1-n} \sum_{k=0}^{\lfloor n/2\rfloor} \frac{(-1)^k}{(n-k)!k!} (n-2k)^{n-1} \sin((n-2k)M_e)$$
It was also given that the above series converges for $|e|<0.662743419$.
Now my questions are

*

*How can we derive an infinite series solutions given such transcendental equation to solve?
(I have no idea.)


*How can we find interval of convergence of this infinite series?
(I have tried using Ratio tests etc., to find interval of convergence but they are not that helpful.)
 A: I shall follow the book [KT].

How can we derive an infinite series solutions given such transcendental equation to solve?

In Subsection 3.3.1 they look for a solution of Kepler’s equation $E-e\sin E=M$ in a form of a series of powers of the eccentricity $e$ with the coefficients dependent of $M$, as follows.
Represent $E$ in a standard form of a Maclaurin  series
$$E=\sum_{k=0}^\infty a_k(M)e^k,\mbox{ where }a_k=\frac{1}{k!}\frac{d^k E}{de^k}{\huge|}_{e=0}.\tag{3.21}$$
Recall that $M$ is considered to be a real parameter. When $e=0$ we have $E=M$, so $a_0=M.$ The remaining coefficients 
$a_k$ periodically depend on $M$, so (3.21) usually is written as 
$$E-M=\sum_{k=0}^\infty a_k(M)e^k.$$
Differentiating Kepler’s equation, we obtain $(1-e \cos E) dE-\sin E de = 0,$ so 
$$\frac {dE}{de}=\frac{\sin E}{1-e\cos E}.$$
Since $M$ is considered as a parameter, we have $dM=0.$
Calculate the second derivative.
$$\frac {d^2E}{de^2}=\left(\frac{\partial}{\partial e}+\frac {dE}{de}\frac{\partial}{\partial E}\right)
\frac {\sin E}{1-e \cos E} =\frac{-5e\sin E + 4 \sin 2E - e \sin 3E}{4(1-e \cos E)^3}.$$
Assume that 
$$\frac{d^k E}{de^k} =\frac{\Phi_k(e,E)}{4^{k-1}(1-e \cos E)^{2k-1}},\tag{3.24}$$
where $\Phi_k$ is a Fourier sine polynomial of order at most $2k-1$, whose coefficients 
are polynomials of $e$ of degree at most $k-1$ with integer coefficients. This hold for $k = 1, 2$:
$$\Phi_1 = \sin E,\,\Phi_2 =  -5e \sin E + 4 \sin 2E - e \sin 3E.$$
The assumption is proved by induction taking into account a recurrence 
$$\Phi_{k+1} = 4(2k-1)(\cos E-e)\Phi_k + (4\sin E-2e\sin 2E)\frac{\partial\Phi_k}{\partial E}+ 
(4 + 2e^2-8e\cos E + 2e^2\cos 2E) \frac{\partial\Phi_k}{\partial e},\tag{3.25}$$
which is can easily derived differentiating (3.24).
Formula (3.25) allows to find $\Phi_k$ up to any required order $k$. Then it remains to put 
$$a_k = \frac 1{4^{k-1}k!}\Phi_k(0,M).$$

How can we find interval of convergence of this infinite series?

They wrote that they do not provide the complete proofs because the latter are too hard. 
As I understood, in Subsection 3.8.2 they consider the series $E(e)$ as a function of a complex variable  $e$
and look for its singularities. The essential those turn out to be solutions of Equation (3.166) 
$1- E\cos e=0$. It follows 
$$e=\frac{E-M}{\sin E}=\frac 1{\cos E}.$$
$\newcommand{\ch}{\operatorname{ch}}$
$\newcommand{\sh}{\operatorname{sh}}$
$\newcommand{\cth}{\operatorname{cth}}$
Let $E=u+iv$. It follows 
$$(u-M) \cos u \ch v + v \sin u \sh v = \sin u \ch v,$$
$$- (u- M) \sin u \sh v + v \cos u \ch v = \cos u \sh v,$$
$$|e| = (\ch^2 v -\sin^2 u)^{-1/2}.\tag{3.180}$$
When $\sin u = 0 \Leftrightarrow u = k\pi$, the first two equations (3.180) transform to
$$ (u-M) \ch v = 0,\, v \ch v = \sh v \Rightarrow v = 0,\, M = u = k\pi,\, |e| = 1.$$
When $\cos u = 0 \Leftrightarrow u = 2k\pi\pm \pi/2$, the first two equations (3.180) transform to
$$v \sh v = \ch v,\, (u-M) \sh v = 0 =0 \Rightarrow u = M = 2k\pi\pm \pi/2.$$
In this case $v = 1.199678640$ is a root of the equation $v = \cth v$, $|e|= 1/\sh v = 0.662743419.$
Thus, the convergence radius $R(M)$ of the expansions with respect to the powers of the eccentricity
is equal to $1$ when $M = k\pi$ and to $R_0=R(\pi/2)=0.662743419$ when $M=2k\pi\pm \pi/2$.
It can be shown that $R_0\le R(M)\le 1$. The number $R_0$ is called Laplace limit. When $|e|<R_0$
then the series of powers of the eccentricity converge for all $M$. When $|e|>R_0$, the series 
diverge for some values of $M$. 
References
[KT] K.V. Kholshevnikov, V.B. Titov, Two-body problem, SPb. State University,  Saint-Petersburg, 2007. (The book is in Russian but you can look at the equations). 
A: Part 1 
The solution can be obtained by the Lagrange inversion theorem [1].
The equation $E - e \sin E = M_e$ is written as
$$\frac{E - M_e}{\sin E} = e.$$
Let $f(E) = \frac{E - M_e}{\sin E}$. 
Clearly, $f$ is analytic at $M_e$, $f(M_e) = 0$, and $f'(M_e) = \frac{1}{\sin M_e} \ne 0$. 
By the Lagrange inversion theorem, we have
$$E = M_e + \sum_{n=1}^\infty b_n e^n$$
where
$$b_n = \frac{1}{n!} \lim_{w\to M_e} \frac{\mathrm{d}^{n-1}}{\mathrm{d} w^{n-1}}
(\sin w)^n.$$
Clearly, $a_0 = 0$, $a_1 = b_1 = \sin M_e$. Let us prove that $a_n = b_n$ for $n \ge 2$. 
It suffices to prove that, for $n\ge 2$,
$$2^{1-n} \sum_{k=0}^{\mathrm{floor}(n/2)} \binom{n}{k} (-1)^k (n-2k)^{n-1}\sin ((n-2k)w)
= \frac{\mathrm{d}^{n-1}}{\mathrm{d} w^{n-1}}(\sin w)^n.$$
If $n$ is odd, by using the identity [2]
$$\sin^n w = \frac{2}{2^n}\sum_{k=0}^{(n-1)/2} (-1)^{\frac{n-1}{2}-k}\binom{n}{k}\sin ((n-2k)w),$$
we have
\begin{align}
\frac{\mathrm{d}^{n-1}}{\mathrm{d} w^{n-1}}(\sin w)^n
&= \frac{2}{2^n}\sum_{k=0}^{(n-1)/2} (-1)^{\frac{n-1}{2}-k}\binom{n}{k}
(n-2k)^{n-1}(-1)^{(n-1)/2}\sin ((n-2k)w)\\
&= \frac{2}{2^n}\sum_{k=0}^{(n-1)/2} (-1)^k\binom{n}{k}
(n-2k)^{n-1}\sin ((n-2k)w).
\end{align}
The desired result follows.
If $n$ is even, by using the identity [2]
$$\sin^n w = \frac{1}{2^n}\binom{n}{n/2} + \frac{2}{2^n}\sum_{k=0}^{n/2 - 1}(-1)^{n/2-k}\binom{n}{k}\cos ((n-2k)w),$$
we have
\begin{align}
\frac{\mathrm{d}^{n-1}}{\mathrm{d} w^{n-1}}(\sin w)^n
&= \frac{2}{2^n}\sum_{k=0}^{n/2 - 1}(-1)^{n/2-k}\binom{n}{k}(n-2k)^{n-1}(-1)^{n/2} \sin ((n-2k)w)\\
&= \frac{2}{2^n}\sum_{k=0}^{n/2 - 1}(-1)^k \binom{n}{k}(n-2k)^{n-1}\sin ((n-2k)w).
\end{align}
The desired result follows.
We are done.
$\phantom{2}$
Part 2 
By root test, the series converges if
$$e < \frac{1}{\limsup_{n\to \infty} \sqrt[n]{|a_n|}}.$$
We have
\begin{align}
\sqrt[n]{|a_n|} &= \left|\frac{1}{n!} 2^{1-n} \sum_{k=0}^{\mathrm{floor}(n/2)} \binom{n}{k} (-1)^k (n-2k)^{n-1}\sin ((n-2k)M_e)\right|^{1/n}\\[5pt]
&\le \left(\frac{1}{n!} 2^{1-n} \sum_{k=0}^{\mathrm{floor}(n/2)} \binom{n}{k} (n-2k)^{n-1}\right)^{1/n}\\[5pt]
&= \left(\frac{1}{n!} 2^{1-n}n^{n-1}\right)^{1/n}\left(\sum_{k=0}^{\mathrm{floor}(n/2)} \binom{n}{k} \left(1-\frac{2k}{n}\right)^{n-1}\right)^{1/n}.
\end{align}
By using $\lim_{n\to\infty} (\frac{1}{n!} 2^{1-n}n^{n-1})^{1/n} = \frac{1}{2}\mathsf{e}$, we have
$$\frac{1}{\limsup_{n\to \infty} \sqrt[n]{|a_n|}} \ge \liminf_{n\to \infty} \frac{2}{\mathsf{e}}\left(\sum_{k=0}^{\mathrm{floor}(n/2)} \binom{n}{k} \left(1-\frac{2k}{n}\right)^{n-1}\right)^{-1/n} \triangleq \lambda.$$
Then, the series converges if
$$e < \lambda.$$
I $\color{blue}{\textrm{GUESS}}$ (note: here $\lim_{n\to \infty}$ rather than $\liminf_{n\to \infty}$)
$$\lim_{n\to \infty} \underbrace{\frac{2}{\mathsf{e}}\left(\sum_{k=0}^{\mathrm{floor}(n/2)} \binom{n}{k} \left(1-\frac{2k}{n}\right)^{n-1}\right)^{-1/n}}
_{B_n}
= 0.662743419... $$
I did some numerical experiments by Maple. For example, it seems $B_n$ is non-increasing; When $n=1000$ (Maple can not easily evaluate $B_n$ for larger $n$),
$$\frac{2}{\mathsf{e}}\left(\sum_{k=0}^{\mathrm{floor}(n/2)} \binom{n}{k} \left(1-\frac{2k}{n}\right)^{n-1}\right)^{-1/n} = 0.6627434531...$$
Reference
[1] https://en.wikipedia.org/wiki/Lagrange_inversion_theorem
[2] https://en.wikipedia.org/wiki/List_of_trigonometric_identities
