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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 $1- (1+\delta)^k (1-\delta)^{n-k}$.

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    $\begingroup$ Because of the exponential growth, for each fixed $\delta \in (0, \frac{1}{4})$ the sum is roughly $ ( (1+\delta)^2 / 2 )^{n+o(1)} $. But I guess what you you want is some asymptotics uniform both in $\delta$ and in $n$. $\endgroup$ Jul 8, 2017 at 3:35
  • $\begingroup$ How do you get $((1+\delta)^2/2)^n$ ? It seems to be too small... $\endgroup$
    – vkonton
    Jul 8, 2017 at 4:04

1 Answer 1

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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 $n.$

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  • $\begingroup$ I drop the absolute value and let $k$ go from $0$ to $n$ not to $2n$. $\endgroup$
    – vkonton
    Jul 8, 2017 at 3:51
  • $\begingroup$ @vkonton ah, ok. But then Series[] returns a perfectly reasonable power series. $\endgroup$
    – Igor Rivin
    Jul 8, 2017 at 11:25
  • $\begingroup$ I just don't know how to lower bound the difference $\left((\delta +1) \left(1-\delta ^2\right)^n \, _2F_1\left(1,1-n;n+2;\frac{\delta +1}{\delta -1}\right)+(\delta -1) \, _2F_1(1,1-n;n+2;-1)\right)$ that appears. $\endgroup$
    – vkonton
    Jul 8, 2017 at 16:33
  • $\begingroup$ @vkonton See the addendum. $\endgroup$
    – Igor Rivin
    Jul 8, 2017 at 17:12
  • $\begingroup$ You're right but since you drop $\binom{2n}{k}$ you get a lower bound $\Omega( n^2 2^{-2n} \delta)$ which is tiny. $\endgroup$
    – vkonton
    Jul 8, 2017 at 17:37

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