Monotonicity of a partial binomial sum I'm wondering when does the following partial sum monotonically increase or decrease in $n$:
$$f_x(n)=\sum^{\lfloor\frac{n}{2}\rfloor}_{k=0}{n \choose k}x^k(1-x)^{n-k}.$$
In theory, if $x<\frac{1}{2}$, the probability converges to 1. On the other hand, if $x>\frac{1}{2}$, it converges to zero. And when $x=\frac{1}{2}$, it converges to $\frac{1}{2}$. 
I initially tried to show that the convergence is monotonically made, meaning that $f_x(n)$ is monotone in $n$. However, I tried several numbers for $x$ and found that it is actually not. 
For example, when $x=0.49$, we have $f_{0.49}(2)\simeq0.7599$, $f_{0.49}(20)\simeq0.6229$, $f_{0.49}(10000)\simeq0.9778$. A similar example can be found for $x>\frac{1}{2}$. 
Is there any intuition behind this non-monotonicity? or is there any increment in $n$ that the monotonicity is guaranteed? 

After I saw the answer from joriki, I only focused on odd $n$ cases and could show the monotonicity. I'll leave my reasoning here for a future reference. Thanks for your comment, joriki. 
Let $n=2m+1$, $m\geq 0$. I will show the sign of $f_p(2m+3)-f_p(2m+1)$ depends on whether $p>\frac{1}{2}$ or $p<\frac{1}{2}$. 
Let $X_{n,p}\sim Binomial(n,p)$. The associated cdf is given by $$F_p(n,k)=P[X_{n,p}\leq k]=\sum_{i=0}^{k}{n\choose i}p^i(1-p)^{n-i}.$$ 
Before begin, note that $f_p(2m+1)=F_p(2m+1,m)$. From the cdf, we have \begin{align}F_p(2m+3,m+1)=&P[X_{2m+3,p}\leq m+1]\\=&P[X_{2m+3,p}\leq m+1|X_{2m+1,p}<m]P[X_{2m+1,p}<m]\\&+P[X_{2m+3,p}\leq m+1|X_{2m+1,p}=m]P[X_{2m+1,p}=m]\\&+P[X_{2m+3,p}\leq m+1|X_{2m+1,p}=m+1]P[X_{2m+1,p}=m+1]\\
=&1\cdot F_{p}(2m+1,m-1)+(2p(1-p)+(1-p)^2){2m+1\choose m}p^m(1-p)^{m+1}\\
&+(1-p)^2{2m+1\choose m+1}p^{m+1}(1-p)^m.
\end{align}
Here, note that $F_p(2m+1,m-1)=F_p(2m+1,m)-{2m+1\choose m}p^m(1-p)^{m+1}$. Plugging this in, the expression after the last equality is the same as
$$F_p(2m+1,m)+(1-2p){2m+1\choose m}p^{m+1}(1-p)^{m+1}.$$
Thus, we have $$F_p(2m+3,m+1)=F_p(2m+1,m)+(1-2p){2m+1\choose m}p^{m+1}(1-p)^{m+1}$$
in which the sign of the last term is determined by whether it is $p>\frac{1}{2}$ or $p<\frac{1}{2}.$
