the radius of convergence of a power series Assume that $a$, and $b$ are fixed complex numbers with $a\neq{0}$. Prove that the radius of convergence of the power series $$1+az+a(a-2b){\frac{z^2}{2!}}+a{(a-3b)^2}{\frac{z^3}{3!}}+\cdots$$ is $1/e|b|.$ I tried working it out using the lim sup of the $n$-th roots, and the ratio test but cannot seem to get anywhere! Here is a sketch of my approach.
I am treating the case $b\neq 0$. Let the $n$-th coefficient be $A_n$. If I assume that $$\lim_{n\rightarrow{\infty}}{\frac{n}{(n!)^{1/n}}}=e,$$ an assertion that is an immediate consequence of Stirling's formula, then I can write $$A_n = \frac{a(a-nb)^{n-1}}{n!}=-{\frac{a(n^{n-1})(b^{n-1})(1-{\frac{a}{bn}})^{n-1}}{n!}}.$$ Thus, $$|A_n|^{1/n}={\frac{|a|^{1/n}n(n)^{-1/n}|b||b|^{-1/n}|1-(a/bn)||1-a/bn|^{1/n}}{(n!)^{1/n}}}.$$ Now taking limits as $n\rightarrow{\infty},$ the right hand side converges to $|b|\lim_{n\rightarrow{\infty}}{\frac{n}{n^{1/n}}}=|b|e$, which is what we want. I feel a little foolish having to engage in the above messy calculation! Am I missing the forest for the trees and an easy answer? Wouldn't be the first time!
 A: (amended)
The correct rendering of the series is
$$
\eqalign{
  & F(z) = 1 + az + a\left( {a - 2b} \right){{z^{\,2} } \over {2!}} + a\left( {a - 3b} \right)^{\,2} {{z^{\,3} } \over {3!}} +  \cdots  =   \cr 
  &  = 1 + a\left( {z + \left( {a - 2b} \right){{z^{\,2} } \over {2!}} + \left( {a - 3b} \right)^{\,2} {{z^{\,3} } \over {3!}} +  \cdots } \right) =   \cr 
  &  = 1 + a\left( {\sum\limits_{1\, \le \,k} {\left( {a - kb} \right)^{\,k - 1} {{z^{\,k} } \over {k!}}} } \right) =   \cr 
  &  = 1 - a\left( {\sum\limits_{1\, \le \,k} {\left( { - 1} \right)^{\,k} a^{\,k - 1} \left( {{{kb} \over a} - 1} \right)^{\,k - 1} {{z^{\,k} } \over {k!}}} } \right) =   \cr 
  &  = 1 + az\left( {\sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} a^{\,k} \left( {{{\left( {k + 1} \right)b} \over a} - 1} \right)^{\,k} {{z^{\,k} } \over {\left( {k + 1} \right)!}}} } \right) \cr} 
$$
Now, note that , for large $k$, we can write
$$
\eqalign{
  & F(z) = 1 + a\left( {\sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} a^{\,k} \left( {{{\left( {k + 1} \right)b} \over a} - 1} \right)^{\,k}
 {{z^{\,k + 1} } \over {\left( {k + 1} \right)!}}} } \right) =   \cr 
  &  \approx 1 + a\sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} a^{\,k} \left( {{{\left( {k + 1} \right)b} \over a}} \right)^{\,k}
 {{z^{\,k + 1} } \over {\left( {k + 1} \right)!}}}  =   \cr 
  &  = 1 + a\sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( {k + 1} \right)^{\,k} {{\left( {bz} \right)^{\,k} } \over {\left( {k + 1} \right)!}}}
  = 1 + a\,W_{\,0} (bz) \cr} 
$$
that is, that $F(z)$ is asymptotic to the [Lambert Function][1] $W_0(bz)$
whose radius of convergence is in fact $|bz|=1/e$.
Note: I am not a "theorist", and I am conscious that the above
is not a rigorous definition, but the substance is there.  I hope you or other readers can put it
in rigorous terms.
