Some Inequalities and Asymptotics for a Weighted Alternate Binomial

We establish strict inequality bounds for the binomial sums ∑n i=0 ( n i ) (−1)i 2i+1 and prove the asymptotic result: n ∑ i=0 ( n i ) (−1)i 2i + 1 ∼ √ π 2 1 √ 2n + 1 , as n→ ∞.


Introduction
We are interested in studying the sums n i=0 n i (−1) i 1 2i+1 for all n ∈ N. The sums are shown to be positive and strictly decreasing with n, having limit zero as n tends to infinity.The main results are bounds for these sums as well as their asymptotic behaviour.

Main Results
Denote a n = n i=0 n i (−1) i 2i+1 .We use the notations exp(x) = e x and x ∧ y = x y in order to make some formulas easier to be read.
Proposition 1 For all n ∈ N, n i=0 n i and the result follows from Propositions 1 and 2.
Summing the following inequalities we establish the upper bound.We have used the famous formula 2 to obtain the first inequality.See Nicholas and Yates (1950).
For sequences x n and y n we write x n ∼ y n when lim  4, No. 3;2012 We can improve these bounds as it is shown by the following theorems and propositions: Proof.Letting m = = 0, since, for sufficiently large y, Note that the upper bounds given in Theorem 2 or in the proof of Proposition 4 may not improve the approximation of a n , relative to that of Theorem 1, for small n.
Noting that 0 < z < 1, follow the steps at the end of theorem's 1 proof.
Propositions 4 and 5 tell us that there exists the limit lim n→∞ b n and that it is equal to √ π/2.

Final Remarks
We observe that the bounds can be improved by dividing the interval (0, π/2) in more than two pieces as we have done here and the most direct approach seems to be piecewise linearly approximating the cosine function both from above and bellow except at a neighborhood of the origin where it is approximated by Taylor's formula.
Although this procedure will improve the bounds for small n, of course it will not change the asymptotic result.We remark that from the bounds for a n obtained in this work we can derive bounds for Catalan's numbers and from the asymptotic result we can derive Wallis product formula.See (Bustoz & Ismail, 1986;Everett, 1970;Paule & Pillwein, 2010) and references therein.
1. Denote n and u n the lower and upper bounds given in Theorem 1. Now, letting n → ∞ we have their asymptotic behaviour: Clearly u n ∼ 6π 11 1 √ 2n+1 and, choosing for example μ = √ (2n 1) ln 2 − 1/n, we have n