Is there easier way to calculate the limit of this function? $$
\lim_{K\rightarrow\infty}\frac{(1-\epsilon)^K}{1+(1-\epsilon)^K}\frac{\sum_{i=1}^{\frac{K-1}{2}}\left(\begin{array}{l}
         K \\
         i \end{array}\right)\left[\left(\frac{2\epsilon-\epsilon^2}{(1-\epsilon)^2}\right)^i-\left(\frac{\epsilon}{1-\epsilon}\right)^i\right]}{(1-\epsilon)^K\left[1+\sum_{i=1}^{\frac{K-1}{2}}\left(\begin{array}{l}
         K \\
         i \end{array}\right)\left(\frac{2\epsilon-\epsilon^2}{(1-\epsilon)^2}\right)^i\right]+\left[1+\sum_{i=1}^{\frac{K-1}{2}}\left(\begin{array}{l}
         K \\
         i \end{array}\right)\left(\frac{\epsilon}{1-\epsilon}\right)^i\right]}=?
$$
where $1>\epsilon>0$. I wonder especially when $\epsilon\rightarrow 0$ but $\epsilon\neq 0$
I even dont know if this is a difficult or an easy question for a mathematician. If you could comment on this matter, I will be happy.
 A: An idea which leads directly to the solution is to write the sums involved as (multiples of) probabilities of events involving some binomial random variables $X_K$ and $Y_K$ with respective parameters $(K,\eta)$ and $(K,\epsilon)$, where $\eta=2\epsilon-\epsilon^2$. To wit, the ratio of interest $R_K$ can be rewritten as
$$
R_K=\frac{(1-\epsilon)^K}{1+(1-\epsilon)^K}\cdot\frac{A_K-B_K}{(1-\epsilon)^K\cdot A_K+B_K},
$$
with
$$
A_K=\sum_{i=0}^{\frac{K-1}{2}}{K\choose i}\left(\frac{\eta}{1-\eta}\right)^i=(1-\eta)^{-K}\cdot\mathrm P(X_K\leqslant\tfrac12(K-1)),
$$
and
$$
B_K=\sum_{i=0}^{\frac{K-1}{2}}{K\choose i}\left(\frac{\epsilon}{1-\epsilon}\right)^i=(1-\epsilon)^{-K}\cdot\mathrm P(Y_K\leqslant\tfrac12(K-1)).
$$
If $\eta\lt\frac12$, then $\epsilon\lt\frac12$ and, by the (weak) law of large numbers, $\mathrm P(X_K\leqslant\tfrac12(K-1))\to1$ and $\mathrm P(Y_K\leqslant\tfrac12(K-1))\to1$ when $K\to\infty$. 
Thus, 
$$
(1-\epsilon)^{2K}\cdot A_K=(1-\eta)^{K}\cdot A_K\to1,\qquad
(1-\epsilon)^{K}\cdot B_K\to1.
$$
Finally, for every $\epsilon$ such that $\eta=2\epsilon-\epsilon^2\lt\frac12$, that is, for every $\epsilon\lt1-\frac{\sqrt2}2=0.393$,
$$
\lim\limits_{K\to\infty}R_K=\tfrac12.
$$
