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.

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
$$\Delta^n f(x)=\frac{2\times4\times\cdots\times (2n)}
{(2x+1)(2x+3)\cdots(2x+2n+1)}.$$
