# Combinatorial identity related to the volume of a ball in $\mathbb{R}^{2k+1}$

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

$$\sum_{i=0}^k {k \choose i} \frac{(-1)^i}{2i+1} = \frac{(k!)^2 2^{2k}}{(2k+1)!}$$

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## 2 Answers

Actually, we can derive the recurrence relation for the sum without going through integrals:

$$\begin{align} 2k(I_k-I_{k-1}) &= 2k\sum_{i=0}^k \left(\binom ki-\binom{k-1}i\right)\frac{(-1)^i}{2i+1} \\ &= 2k\sum_{i=0}^k \binom ki\left(1-\frac{k-i}k\right)\frac{(-1)^i}{2i+1} \\ &= 2k\sum_{i=0}^k \binom ki\frac ik\frac{(-1)^i}{2i+1} \\ &= \sum_{i=0}^k \binom ki(-1)^i\frac{2i}{2i+1} \\ &= \sum_{i=0}^k \binom ki(-1)^i\left(1-\frac{1}{2i+1}\right) \\ &= -\sum_{i=0}^k \binom ki(-1)^i\frac{1}{2i+1} \\ &= -I_k\;, \end{align}$$

where I made use of the fact that the alternating sum of $\binom ki$ over $i$ vanishes.

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Thanks! This is the kind of thing I was looking for. – spin Oct 19 '12 at 12:10

You might not be happy about this answer, but I'd actually compute that sum exactly in the way in which I suspect you arrived at it:

$$\begin{align} \sum_{i=0}^k {k \choose i} \frac{(-1)^i}{2i+1} &= \int_0^1\sum_{i=0}^k {k \choose i} (-1)^iq^{2i}\mathrm dq \\ &= \int_0^1(1-q^2)^k\mathrm dq \\ &=\int_0^{\pi/2}(1-\sin^2\theta)^k\cos\theta\,\mathrm d\theta \\ &=\int_0^{\pi/2}\cos^{2k+1}\theta\,\mathrm d\theta \\ &=:I_k \;, \end{align}$$

and then integration by parts yields the recurrence relation

$$I_k=\int_0^{\pi/2}\cos^{2k}\theta\cos\theta\,\mathrm d\theta=2k\int_0^{\pi/2}\cos^{2k-1}\theta\sin^2\theta\,\mathrm d\theta=2k(I_{k-1}-I_k)$$

and thus

$$I_k=\frac{2k}{2k+1}I_{k-1}$$

with $I_0=1$, whose closed-form solution is readily seen to be your right-hand side.

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