# 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.

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@anon: That's what I thought too, but he says $\epsilon \ne 0$... –  rubik Jul 7 '12 at 18:19
yes it cannot be zero but can be a very small positive real number. –  Seyhmus Güngören Jul 7 '12 at 18:19
Oh, sorry, I thought it was $\lim\limits_{\epsilon\to0^+}$. Then I would just "reverse-use" the binomial theorem. –  anon Jul 7 '12 at 18:20
yes I saw the theorem: $(1+x)^n=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right)x^k$ –  Seyhmus Güngören Jul 7 '12 at 18:31
@SeyhmusGüngören $p^i q^{n-i}=q^n(p/q)^i$ and the first multiplier does not depend on $i$ so... –  Andrew Jul 7 '12 at 20:18

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 $\epsilon\lt1-\frac{\sqrt2}2=0.393$, $$\lim\limits_{K\to\infty}R_K=\tfrac12.$$
Thanks alot for the solution. I have one more question named "maximum of the function at limit" in this question the first part is always $1$ if $\theta <0.5$. but the second part is diverging from 1 if $\theta<0.5$ but very close to $0.5$. For example if $\theta=0.49$ then I have $(1-t)^K f=0.7364$ and the first part is $1$ and the difference is $0.2636$. I took $K=999$. I guess K is not enough since $\theta$ is very close to $0.5$ although less than that of. Then the answer for my other question should be $0$? then it seems strange to me because this question was general of the other one.. –  Seyhmus Güngören Jul 8 '12 at 14:50