Formula for $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\cdots$ I was doing an exercise about the integral $$ I_{n} = \int_{0}^{1} (1-x^{2})^{n}\, \mathrm{d}x $$ and I tried two approaches. On the one hand, by substituting $x = \sin u $ I found that $$ I_{n} = \frac{2}{3} × \frac{4}{5} × \cdots × \frac{2n}{2n+1} = \left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\cdots\left(1-\frac{1}{2n+1}\right).$$
On the other hand, the binomial expansion formula gave me $$I_{n} = \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{k}}{2k+1}.$$
So I find that $$ \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{k}}{2k+1} = \left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\cdots\left(1-\frac{1}{2n+1}\right),$$
but I want to understand this identity without integration. Is there a direct way to prove this? Perhaps by induction or generating polynomials or something.
I tried induction, and found that proving the identity is equivalent to proving $$ \sum_{k=0}^{n} \binom{n}{k}(-1)^{k} \left( \frac{1}{2n+3} - \frac{1}{2k+3} \right) = 0 $$
but here I get stuck.
 A: An expression
$$\sum_{k=0}^n (-1)^k\binom{n}k f(a+k)$$
indicates an $n$-fold differencing operation. Here you have $f(x)=1/(2x+1)$
and $a=0$.
Define
$$\Delta f(x)=f(x)-f(x+1),$$
$$\Delta^2 f(x)=\Delta f(x)-\Delta f(x+1)=f(x)-2f(x+1)+f(x+2)$$
etc. Then
$$\Delta^n f(x)=\sum_{k=0}^n (-1)^k\binom{n}k f(x+k).$$
In our example $f(x)=1/(2x+1)$. So
$$\Delta f(x)=\frac1{2x+1}-\frac1{2x+3}=\frac2{(2x+1)(2x+3)}$$
$$\Delta^2 f(x)=\frac2{(2x+1)(2x+3)}-\frac2{(2x+3)(2x+5)}
=\frac{2\times 4}{(2x+1)(2x+3)(2x+5)}$$
etc. By induction
$$\Delta^n f(x)=\frac{2\times4\times\cdots\times (2n)}
{(2x+1)(2x+3)\cdots(2x+2n+1)}.$$
A: Let us consider the function
$$ f_n(z) = \prod_{k=1}^{n}\frac{2k}{2k+1+z}. \tag{1}$$
That is a meromorphic function with simple poles at $z=-3,-5,-7,\ldots,-(2n+1)$ and the residue at $z=-(2k+1)$ is given by
$$ \lim_{z\to -(2k+1)} (2k+1+z)\,f_n(z) = \prod_{\substack{j=1 \\j\neq k}}^{n}\frac{2j}{2j+1-(2k+1)}=\frac{(2n)!!/(2k)}{2^{n-1}(-1)^{k-1}(k-1)!(n-k)!} $$
so the partial fraction decomposition of $f_n(z)$ is the following one:
$$ \prod_{k=1}^{n}\frac{2k}{2k+1+z} = \sum_{k=1}^{n}\frac{(-1)^{k+1}\binom{n}{k}}{z+2k+1}\tag{2}$$
and the claim follows by evaluating $(2)$ at $z=0$.

That gives an interesting equivalent approach to the "usual" approach through integration and the properties of Euler's Beta function:
$$\begin{eqnarray*}\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{2k+1}&=&\int_{0}^{1}\sum_{k=0}^{n}\binom{n}{k}(-1)^k x^{2k}\,dx\\&=&\int_{0}^{1}(1-x^2)^n\,dx\\&=&\frac{1}{2}\int_{0}^{1}z^{-1/2}(1-z)^n\,dz\\&=&\frac{1}{2}\,B\left(\frac{1}{2},n+1\right)\\&=&\frac{\Gamma\left(\frac{1}{2}\right)\,\Gamma(n+1)}{2\,\Gamma\left(n+\frac{3}{2}\right)}=\frac{(2n)!!}{(2n+1)!!}.\end{eqnarray*}\tag{3}$$

This proves the sequence given by $a_k=\frac{1}{2k+1}$ and the sequence given by $b_k=\frac{(2k)!!}{(2k+1)!!}$ are the binomial transform of each other. In particular:
$$ \frac{\pi}{4}=\sum_{k\geq 0}\frac{(-1)^k}{2k+1} = \sum_{k\geq 0}\frac{(2k)!!}{(2k+1)!! 2^{k+1}} = \sum_{k\geq 0}\frac{2^k}{\binom{2k}{k}(4k+2)}\tag{4}$$
provides an acceleration for the Gregory series for $\pi$.

Side note: $(2k)!!$ stands for $2\cdot 4\cdots (2k)$ and $(2k+1)!!$ stands for $1\cdot 3\cdots (2k+1)$.
A: Inductive proof.
Letting $$f(n)=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\ldots\left(1-\frac{1}{2n+1}\right)$$
and $$g(n)=\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{2k+1}$$
our goal is to prove that $f (n) = g (n) $ for all $n $.
The key to the induction is that we also have:

Lemma: For $n>0$:
  $$g(n)=\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{k+1}2k}{2k+1}$$
This is because $\frac{-2k}{2k+1}=\frac{1}{2k+1}-1$ and $\sum (-1)^k\binom{n}{k}=0$.

We'll also use that $n\binom{n-1}{k}=(n-k)\binom{n}{k}$.
Induction step:
Assume:
$$g(n-1)= f(n-1)$$
Then:
$$\begin{align}(2n+1)f(n)&=2nf(n-1)\\&=2ng(n-1)\\&=2n\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{(-1)^{k}}{2k+1}\\
&=2\sum_{k=0}^{n-1}\binom{n}{k}\frac{(-1)^k(n-k)}{2k+1}\\
&=2n\sum_{k=0}^{n-1}\binom{n}{k}\frac{(-1)^k}{2k+1}+2\sum_{k=0}^{n-1}\binom{n}{k}\frac{(-1)^{k+1}k}{2k+1}\\
&=2n\left(g(n)-\binom{n}{n}\frac{(-1)^n}{2n+1}\right)+2\sum_{k=0}^{n-1}\binom{n}{k}\frac{(-1)^{k+1}k}{2k+1}\\
&=2ng(n)+\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{k+1}2k}{2k+1}\\&=(2n+1)g(n)
\end{align}$$
