A monotonicity proof of a sequence summation I meet a problem:
I have a conjecture that the following sequence summation is monotony increasing in N and tends to 1 as N tends to $\infty$
$$\sum_{k=0}^{2N+1}C_{2N+1}^{k}\pi^{k}(1-\pi)^{2N+1-k}\frac{(1-q)\pi^{k}(1-\pi)^{2N+1-k}}{(1-q)\pi^{k}(1-\pi)^{2N+1-k}+q(1-\pi)^{k}\pi^{2N+1-k}}$$
where $0<\pi<0.5,0<q<0.5$ and q is a certain prior probability.
This expression is an expectation in my problem under certain background. The fraction term comes from Bayesian formula.
I try some matlab simulation and this conjecture is right but how to prove it via tricky method?
 A: Let $0<b:=\frac{q}{1-q}<1$ and $0<x:=\frac{\pi}{1-\pi}<1$. So, your sum now takes the form
$$B(N):=\sum_{k=0}^{2N+1}\binom{2N+1}k\frac{\pi^k(1-\pi)^{2N+1-k}}{1+b\cdot x^{2N+1-2k}}.$$
We show monotonicity: $B(N+1)-B(N) > 0$.
Applying Pascal's recurrence twice, we get $\binom{2N+3}k=\binom{2N+1}k+2\binom{2N+1}{k-1}+\binom{2N+1}{k-2}$. After shifting indices, in the last two summations, we gather that
\begin{align} B(N+1)
&=(1-\pi)^2\sum_{k=0}^{2N+1}\binom{2N+1}k\frac{\pi^k(1-\pi)^{2N+1-k}}{1+b\cdot x^2\cdot x^{2N+1-2k}} \\
&+2\pi(1-\pi)\sum_{k=0}^{2N+1}\binom{2N+1}k\frac{\pi^k(1-\pi)^{2N+1-k}}{1+b\cdot x^{2N+1-2k}} \\
&+\pi^2\sum_{k=0}^{2N+1}\binom{2N+1}k\frac{\pi^k(1-\pi)^{2N+1-k}}{1+b\cdot x^{-2}\cdot x^{2N+1-2k}}.
\end{align}
Letting $y:=x^{2N+1-2k}$, the difference $B(N+1)-B(N)$ equals
\begin{align}
&\sum_{k=0}^{2N+1}\binom{2N+1}k\pi^k(1-\pi)^{2N+1-k}\left[
\frac{(1-\pi)^2}{1+bx^2y}+\frac{2\pi(1-\pi)-1}{1+by}+\frac{\pi^2}{1+bx^{-2}y}\right] \\
=&\sum_{k=0}^{2N+1}\binom{2N+1}k\frac{\pi^k(1-\pi)^{2N+1-k}}{(1+x)^2}\left[
\frac1{1+bx^2y}-\frac{1+x^2}{1+by}+\frac{x^2}{1+bx^{-2}y}\right] \\
&=\sum_{k=0}^{2N+1}\binom{2N+1}k\frac{\pi^k(1-\pi)^{2N+1-k}}{(1+x)^2}\left[
\frac{b^2y^2(1+x^2)(1-x^2)^2}{(1+bx^2y)(1+by)(x^2+by)}\right]>0.
\end{align}
We've proven that each quantity inside $[\cdots]$ is term-wise positive, which is stronger than saying the sum $B(N+1)-B(N)>0$.
Since $1+b\cdot x^{2N+1-2k}>1$, it is clear that your sequence
$$B(N)<\sum_{k=0}^{2N+1}\binom{2N+1}k\pi^k(1-\pi)^{2N+1-k}=(\pi+1-\pi)^{2N+1}=1.$$
So, $B(N)$ is increasing and bounded above; hence by the Monotone Convergence Theorem, this sequence has a limit on the real line $\mathbb{R}$. The analysis for $B(N)\rightarrow1$ has already been discussed by Antony Quas. However, in case you wish to see some clues, notice you can estimate the term
$$\frac1{1+b\cdot x^{2N+1-2k}}=\frac{x^{2k}}{x^{2k}+b\cdot x^{2N+1}}\sim
\frac{x^{2k}}{x^{2k}}=1$$
due to the fact that $0<x<1$ and $x^{2N+1}\rightarrow0$, as $N\rightarrow\infty$. Therefore, you estimates $B(N)$ for large $N$ such that
\begin{align} B(N)
&=\sum_{k=0}^{2N+1}\binom{2N+1}k\pi^k(1-\pi)^{2N+1-k}\frac{x^{2k}}{x^{2k}+b\cdot x^{2N+1}} \\
& \sim \sum_{k=0}^{2N+1}\binom{2N+1}k\pi^k(1-\pi)^{2N+1-k} \\
&=(\pi+1-\pi)^{2N+1}=1.
\end{align}
A: So showing this converges to 1 shouldn't be too hard. The fraction is at most 1, and is close to 1 for $k<(1-\epsilon)N$ (for any $\epsilon>0$ and for all large $N$). Since almost all of the mass of the binomial distribution is concentrated around $k\approx (2N+1)\pi$, you can choose an $\epsilon>0$ so that $1-\epsilon>2\pi$. Now for most of the mass, the fraction is close to 1, and so the sum is close to the binomial expansion of $(\pi+(1-\pi))^{2N+1}$. 
