TITLE: Complicated sum with binomial coefficients
QUESTION [11 upvotes]: I know how to prove, that $\frac{1}{2^{n}}\cdot\sum\limits_{k=0}^nC_n^k \cdot
\sqrt{1+2^{2n}v^{2k}(1-v)^{2(n-k)}}$ tends to 2 if n tends to infinity for $v\in (0,\, 1),\ v\neq 1/2$. This can be proved with the use of Dynamical systems reasonings, which are non-trivial and complicated. Is there quite good combinatiorial proof of this fact? How would you solve this problem as it is? Here $C_n^k$ denotes $\frac{n!}{k!(n-k)!}$.

REPLY [8 votes]: Let $$x=\sqrt[4]{4^nv^{2k}(1-v)^{2n-2k}}=(\sqrt{2})^n (\sqrt{v})^{k}(\sqrt{1-v})^{n-k},x>0$$
First, we have following inequality
$$x^2-x+1<\sqrt{x^4+1}<x^2+1,x>0\tag{1}$$ 
because 
$$x^4+1>(x^2-x+1)^2\Longleftrightarrow x(2x^2-3x+2)>0,x>0$$
$$(x^4+1)<(x^2+1)^2\Longleftrightarrow x^2>0$$
use $(1)$, we have
$$1+2^nv^k(1-v)^{n-k}-(\sqrt{2})^n(\sqrt{v})^k(\sqrt{1-v})^{n-k}
<\sqrt{1+4^nv^{2k}(1-v)^{2n-2k}}<1+2^nv^k(1-v)^{n-k}\tag{2}$$
use $(2)$,For one thing,we have 
\begin{align*}\lim_{n\to\infty}\dfrac{\displaystyle\sum_{k=0}^{n}\binom{n}{k}\sqrt{1+4^nv^{2k}(1-v)^{2n-2k}}}{2^n}
&<\lim_{n\to+\infty}\dfrac{\displaystyle\sum_{k=0}^{n}\binom{n}{k}(1+2^nv^{k}(1-v)^{n-k})}{2^n}\\
&=\lim_{n\to+\infty}\dfrac{\displaystyle\sum_{k=0}^{n}\binom{n}{k}+\sum_{k=0}^{n}\binom{n}{k}2^nv^k(1-v)^{n-k}}{2^n}\\
&=\lim_{n\to+\infty}\dfrac{(1+1)^n+2^n(v+1-v)^n}{2^n}\\
&=2
\end{align*}
use $(2)$,on the other hand 
$$\lim_{n\to\infty}\dfrac{\displaystyle\sum_{k=0}^{n}\binom{n}{k}\sqrt{1+4^nv^{2k}(1-v)^{2n-2k}}}{2^n}
>2-\lim_{n\to+\infty}\dfrac{(\sqrt{2v}+\sqrt{2(1-v)})^n}{2^n}=2$$
because Use Cauchy-Schwarz inequality we have
$$\sqrt{v}+\sqrt{1-v}<\sqrt{2(v+1-v)}=\sqrt{2}$$