Lower Bound for Sum.

Question

I want to find a lower bound for the sum \begin{equation} (1/2)^{2n} \sum_{k=0}^{2n} \binom{2n}{k} \left|1- (1+\delta)^k (1-\delta)^{2n-k} \right| \end{equation} where $n>1$, $0<\delta<1/4$.

If I drop the absolute value and let $k$ go from $0$ to $n$ Mathematica is able to deduce a closed form expression \begin{equation} \frac{4^{-n} \binom{2 n}{n+1} \left((\delta +1) \left(1-\delta ^2\right)^n \, _2F_1\left(1,1-n;n+2;\frac{\delta +1}{\delta -1}\right)+(\delta -1) \, _2F_1(1,1-n;n+2;-1)\right)}{1-\delta} \end{equation} where $F_1$ is the Hypergeometric function.

Question: Is there any way to deduce a simpler lower bound of the above expression ? The hard part for me is to prove that \begin{equation} \frac{(\delta +1) \left(1-\delta ^2\right)^n}{1-\delta} \, _2F_1\left(1,1-n;n+2;\frac{\delta +1}{\delta -1}\right)- \, _2F_1(1,1-n;n+2;-1) \geq n\delta/ C \end{equation} where $C>0$ is a constant.

My attempt

The first order Taylor Approximation for $\delta$ around $0$ for the expression $1- (1+\delta)^k (1-\delta)^{2n-k}$ is $2(n-k) \delta$.

Then the sum \begin{equation} (1/2)^{2 n} \sum_{k=0}^{n} \binom{2n}{k} 2 (n-k) \delta = \delta\ 2^{-2 n-1} (n+1) \binom{2 n}{n+1} \asymp \frac{1}{2 \sqrt{\pi }} \sqrt{n} \delta. \end{equation} This is exactly the lower bound I want to get, namely $\Omega(\sqrt{n} \delta)$, but I cannot find a formal argument at moment. My problem is that $2(n-k) \delta$ is not a lower bound for the expression $1- (1+\delta)^k (1-\delta)^{n-k}$.

Answer 1

If you drop the absolute value, the sum equals $0,$ no matter what mathematica says, so this gives you absolutely nothing useful. Otherwise, the term inside absolute value (for very small $\delta$) is positive for $k> n,$ and negative otherwise. If you break up the sum in this way, you get a hypergeometric expression which mathematica can compute the serie of at zero.

ADDENDUM

Defining

dog[n_, x_] := (1/2)^(2 n) Sum[(1 - (1 + x)^k (1 - x)^(2 n - k)), {k, 
0, n - 1}]

And then evaluating dog[n, x] has no hypergeometrics.

Also:

 dog[n_, x_] := (1/2)^(2 n) Sum[Binomial[2 n, k] (1 - (1 + x)^k (1 - x)^(2 n - k)), {k, 0, n - 1}]

Series[dog[n, x], {x, 0, 1}]

gives

$$2^{-2 n} n x \binom{2 n}{n}+O\left(x^2\right).$$

WHich is about $O(\sqrt{n})$ in $n.$