Limit of quotient of two similar series Let $0 < p < 1,0 < q < 1$. Find the limit $L = \mathop {\lim }\limits_{n \to \infty } \left( {1 - p} \right)\frac{{\sum\limits_{m = 0}^\infty  {{{\left( {1 - p} \right)}^m}{{\left( {1 - {{\left( {1 - p} \right)}^{m + 1}}} \right)}^n}{{\left( {1 - q} \right)}^m}} }}{{\sum\limits_{m = 0}^\infty  {{{\left( {1 - {{\left( {1 - p} \right)}^{m + 1}}} \right)}^n}{{\left( {1 - q} \right)}^m}} }}$.
I've managed to get that the limit above is equal to $1 - p\mathop {\lim }\limits_{n \to \infty } \dfrac{{\sum\limits_{i = 0}^n {{{\left( { - 1} \right)}^i}\left( \begin{gathered}
  n \\ 
  i \\ 
\end{gathered}  \right)\dfrac{1}{{1 - \left( {1 - q} \right){{\left( {1 - p} \right)}^{i + 1}}}}\dfrac{{{{\left( {1 - p} \right)}^i}}}{{1 - \left( {1 - q} \right){{\left( {1 - p} \right)}^i}}}} }}{{\sum\limits_{i = 0}^n {{{\left( { - 1} \right)}^i}\left( \begin{gathered}
  n \\ 
  i \\ 
\end{gathered}  \right)\dfrac{{{{\left( {1 - p} \right)}^i}}}{{1 - \left( {1 - q} \right){{\left( {1 - p} \right)}^i}}}} }}$, I'm not sure which one is easier to compute.
EDIT: It's obvious that $0 \leqslant L \leqslant 1 - p$.
 A: The answer is $L=0$.
Indeed, let
$$F_n=q\sum_{m=0}^\infty\left(1-(1-p)^{m+1}\right)^n(1-q)^m$$
Then
$$
F_n-F_{n+1}=q(1-p)\sum_{m=0}^\infty(1-p)^m\left(1-(1-p)^{m+1}\right)^n(1-q)^m
$$
Thus
$$\eqalignno{
A_n~&{\buildrel{\rm def}\over=}~\frac{(1-p)\sum_{m=0}^\infty(1-p)^m\left(1-(1-p)^{m+1}\right)^n(1-q)^m}{\sum_{m=0}^\infty\left(1-(1-p)^{m+1}\right)^n(1-q)^m}
\cr
&=1-\frac{F_{n+1}}{F_n}.& (1)}
$$
Now, for  a given $r\geq0$ we have
$$\eqalign{
\left(1-(1-p)^{r+1}\right)\root{n}\of{q(1-q)^{r}}
&\leq
\root{n}\of{q\sum_{m=0}^\infty\left(1-(1-p)^{m+1}\right)^n(1-q)^m} 
=\root{n}\of{F_n}\cr
&\leq
\root{n}\of{q\sum_{m=0}^\infty (1-q)^m}=1
}
$$
Rcalling that $\lim_{n\to\infty} \root{n}\of{q(1-q)^{r}}=1$, we conclude that:
$$
1-(1-p)^{r+1}\leq\liminf_{n\to\infty}\root{n}\of{F_n}
\leq\limsup_{n\to\infty}\root{n}\of{F_n}\leq 1
$$
But $r$ is arbitrary, so letting $r$ tend to $+\infty$ we get
$$
1\leq\liminf_{n\to\infty}\root{n}\of{F_n}
\leq\limsup_{n\to\infty}\root{n}\of{F_n}\leq 1
$$
Or $\lim_{n\to\infty} \root{n}\of{F_n}=1$. This implies that 
$$
\lim_{n\to\infty}\frac{F_{n+1}}{F_n}=1,\tag{2}
$$
if we know that this limit does exist, and consequently, that
$$
L=\lim_{n\to\infty}A_n=0
$$
according to $(1)$.
Now to see that the limit in $(2)$ does exist, note that
$$
X^2F_n+2XF_{n+1}+F_{n+2}=
q\left(\sum_{m=0}^\infty (1-q)^m\left(1-(1-p)^{m+1}\right)^n(X+1-(1-p)^{m+1})^2\right)\ge0$$
Hence the discriminant of this second degree trinomial is negative. That is
$$F_{n+1}^2\leq F_nF_{n+2}$$
This is equivalent to the fact that $\left(F_{n+1}/F_n\right)_{n\ge1}$ is a monotone increasing sequence, and the limit $\lim\limits_{n\to\infty}F_{n+1}/F_n$ does exist.
