Calculating the volume of a ball in $\mathbb{R}^{2k+1}$ in two different ways gives us the following formula:
$\sum_{i=0}^k {k \choose i} \frac{(-1)^i}{2i+1} = \frac{(k!)^2 2^{2k}}{(2k+1)!}$
Is there a more direct way to prove this identity? I'm interested if there is a more combinatorial or algebraic way to prove this. Given the sum on the left side, how would you find out the formula for it?
Added: This is how I found the identity. The volume of an ball of radius $r$ in $\mathbb{R}^{2k+1}$ is given by the formula $\mathscr{L}^{2k+1}(B(0,r)) = \frac{\pi^k k! 2^{2k+1}}{(2k+1)!}r^{2k+1}$
and in $\mathbb{R}^{2k}$ by the formula
$\mathscr{L}^{2k}(B(0,r)) = \frac{\pi^k}{k!}r^{2k}$
where $\mathscr{L}$ denotes Lebesgue measure. I was wondering if I could prove the formula for $\mathbb{R}^{2k+1}$ using the formula for $\mathbb{R}^{2k}$. With the formula for even dimension we can calculate
\begin{align*} \mathscr{L}^{2k+1}(B(0,r)) &= (\mathscr{L}^{2k} \times \mathscr{L})(B(0,r)) \\ &= \int_{[-r,r]} \mathscr{L}^{2k}(B(0,\sqrt{r^2 - y^2}))d \mathscr{L}(y) \\ &= \frac{\pi^k}{k!} 2 \int_0^r (r^2 - y^2)^k dy \\ &= \frac{\pi^k}{k!} 2r^{2k+1} \sum_{i=0}^k {k \choose i}\frac{(-1)^i}{2i+1} \end{align*}
Now equating the two formulas for $\mathscr{L}^{2k+1}(B(0,r))$ gives
$\sum_{i=0}^k {k \choose i} \frac{(-1)^i}{2i+1} = \frac{(k!)^2 2^{2k}}{(2k+1)!}$