The binomial sum is as follows:
$$\mathcal {L}^K(\theta)= \sum_{i=\lceil{K/2}\rceil}^K \binom{K}{i}\theta^i\left((1-\theta)^{K-i}-\frac{1}{2}(1-\theta)^{-K}(1-2\theta)^{K-i}\right)$$
which can also be written as
$$\mathcal {L}^K(\theta)= B(K/2,K,1-\theta)-\frac{1}{2}B\left(K/2,K,\frac{1-2\theta}{1-\theta}\right)$$ where $B$ is the Binomial c.d.f.
It can be found that as $K\to\infty$,
\begin{equation} \quad\mathcal {L}^\infty(\theta)= \begin{cases} 0, & \mbox{if } \mbox{ $\theta<\frac{1}{3}$} \\ -\frac{1}{2} & \mbox{if } \mbox{ $\frac{1}{3}<\theta<\frac{1}{2}$} \end{cases}\nonumber \end{equation}
My question is however not about it. I want to show that $\mathcal {L}^K(\theta)$ is negative if $K$ is chosen large enough, for every $\theta\in(0,0.5)$.
This question is equivalent to finding that the zero crossing point $\theta_0\in(0,0.5)$ of $\mathcal {L}^K$ tends to $0$ as $K\to\infty$ and for every $\theta>\theta_0$, $\mathcal {L}^K$ is negative. Notice that $K$ is an odd number for all cases.
A mathematica plot confirms that it must be the case. Here is the plot:
But I dont have any idea on how to show it. Does anyone have?