TITLE: A relation between a binomial sum and a trigonometric integral
QUESTION [6 upvotes]: May not be a research-level problem for an expert, but non-trivial for a non-expert: why do we have
  $$ \sum_{k=0}^n \frac{(-1)^k}{2k+1} \binom{n}{k} = \frac12 \int_0^\pi (\sin x)^{2n+1} dx $$
and what is the asymptotic / good lower bound for this as $n$ grows? Thanks!

REPLY [11 votes]: As I hinted in my comment, the $(-1)^k \binom{n}{k}...$ should set off alarm bells. There is for example a basic reciprocity law, that for functions $f, g: \mathbb{N} \to \mathbb{R}$, we have 
$$g(n) = \sum_{k=0}^n (-1)^k \binom{n}{k} f(k) \qquad \text{iff} \qquad f(n) = \sum_{k=0}^n (-1)^k \binom{n}{k} g(k)$$ 
(see Concrete Mathematics by Graham, Knuth, and Patashnik, Trick 3 on page 192; it's an easy exercise using e.g. exponential generating functions). 
Using this, the claimed identity is equivalent to 
$$\frac{1}{2n+1} = \sum_{k=0}^n \binom{n}{k} \frac1{2} \int_0^\pi (-1)^k (\sin x)^{2k+1}\; dx$$ 
which follows easily since the right side is (by the binomial theorem) 
$$\frac1{2} \int_0^\pi \sin(x) (1 - \sin^2 x)^n\; dx = \frac1{2} \int_0^\pi (\cos x)^{2n} \sin x\; dx$$ 
followed by a straightforward integration. 

As noted by Ira Gessel, the closed-form evaluation of $\sum_{k=0}^n (-1)^k \binom{n}{k} \frac1{2k+1}$ can be achieved by any number of methods. In the spirit of finite difference calculus (alluded to in my comment under the question), where the $n^{th}$ falling power 
$$x^{\underline{n}} = x(x-1)\ldots (x-n+1)$$ 
plays the role of $x^n$ in the ordinary continuous calculus, we have $x^{\underline{m+n}} = x^{\underline{m}} (x - m)^{\underline{n}}$, giving the definition $x^{\underline{-m}} = \frac1{(x+1)(x+2)\ldots (x+m)}$, and the difference formula $\Delta x^{\underline{k}} = k x^{\underline{k-1}}$ (the Pascal triangle identity in disguise) holds for all integers $k$. By means of this one quickly computes 
$$\Delta^n \frac1{2x + 1} = \frac{(-1)^n n! 2^n}{(2x+1)(2x + 3)\ldots (2(x+n) + 1)}$$ 
and then my comment which says $(-1)^n (\Delta^n f)(0) = \sum_{k=0}^n (-1)^k \binom{n}{k} f(k)$ leads to the evaluation 
$$\sum_{k=0}^n (-1)^k \binom{n}{k} \frac1{2k+1} = \frac{n! 2^n}{1 \cdot 3 \cdot \ldots \cdot (2n+1)} = \frac{(n!)^2 2^{2n}}{(2n+1)!}$$ 
as Ira Gessel and T. Amdeberhan were saying.