During some work (i.e. trying to prove series expansion of a certain function) I stumbled upon this identity, which I can't however prove:
$$
\sum\limits_{i=0}^k \frac{(-1)^i}{(n+2i)}\cdot \frac{1}{i!(k-i)!} = \frac{2^k}{n(n+2)(n+4)\ldots(n+2k)}.
$$
Here $n$ is just a fixed positive integer (a dimension of a space, but it's not important). It works in low cases, but all I can do is simply rearrange it term by term, which obviously isn't optimal for general case. I tried induction and some combinatorics tricks, but I'm not good enough at using them.
The reason for Gamma function in the title is that the Maple simply evaluates both sides as
$$
\frac{\Gamma\left(\frac{n}{2}\right)}{2 \cdot \Gamma \left(\frac{n}{2}+k+1\right)}.
$$
I can see it at the right hand side, but the relation of the sum on the LHS to the Gamma function seems out of the blue to me. Does anybody have any hints or ideas for that identity?
Best Answer
You want to multiply both sides by $k!$. Then the LHS becomes
$$\sum_{i=0}^k \frac{(-1)^i}{n + 2i} {k \choose i}$$
which is a bit easier to understand. Starting from the binomial expansion $(1 - x)^k = \sum_{i=0}^k (-1)^i {k \choose i} x^i$ we see that the LHS is
$$I(n, k) = \int_0^1 x^{n-1} (1 - x^2)^k \, dx.$$
This is a variant of the beta function integral. It can be converted into a beta function integral on the nose using the substitution $u = x^2$, which gives
$$I(n, k) = \frac{1}{2} \int_0^1 u^{ \frac{n}{2} - 1 } (1 - u)^k \, du = \frac{1}{2} B \left( \frac{n}{2}, k+1 \right) = \frac{1}{2} \frac{\Gamma \left( \frac{n}{2} \right) \Gamma(k+1)}{\Gamma \left( \frac{n}{2} + k + 1 \right)}$$
and after dividing by $k! = \Gamma(k+1)$ we get Maple's answer.
Edit: For a direct proof starting from the RHS, replace $n$ with a complex variable $z$. Then the LHS is the partial fraction decomposition of the RHS; that is, we in fact have
$$\sum_{i=0}^k \frac{(-1)^i}{i! (k-i)! (z + 2i)} = \frac{2^k}{z(z + 2) \dots (z + 2k)}$$
for complex $z \neq -2i$. To see this it suffices to compute
$$\lim_{z \to -2i} \frac{2^k (z + 2i)}{z(z + 2) \dots (z + 2k)} = \frac{2^k}{(-2i)(-2i + 2) \dots (2k - 2i)} = \frac{(-1)^i}{i! (k-i)!}.$$
Edit #2: Another direct proof, this time starting from the LHS. If $a_n$ is a sequence, write $(\Delta a)_n = a_{n+1} - a_n$ for its forward finite difference. Then we have the general identity
$$(\Delta^k a)_0 = \sum_{i=0}^k (-1)^{k-i} {k \choose i} a_i.$$
Setting $a_i = \frac{1}{n + 2i}$ gives the LHS times $(-1)^k k!$. Now we can argue by induction on $k$: we have
$$(\Delta a)_i = \frac{1}{n + 2i + 2} - \frac{1}{n + 2i} = \frac{-2}{(n + 2i)(n + 2i + 2)}$$ $$(\Delta^2 a)_i = \frac{-2}{(n + 2i + 2)(n + 2i + 4)} - \frac{-2}{(n + 2i)(n + 2i + 2)} = \frac{2^2 2!}{(n + 2i)(n + 2i + 2)(n + 2i + 4)}$$
and in general by induction
$$(\Delta^k a)_i = \frac{(-1)^k 2^k k!}{(n + 2i)(n + 2i + 2) \dots (n + 2i + 2k)}.$$
Setting $i = 0$ then gives the desired identity. As in the previous argument $n$ can be a complex number $z \neq -2i$.
The $2$s are in some sense a red herring. Really this identity is a mild variant of the identity
$$\sum_{i=0}^k \frac{(-1)^i}{z + i} {k \choose i} = \frac{k!}{z(z + 1) \dots (z + k)}.$$
which can be proven using any of the above approaches. Then substitute $z \mapsto \frac{z}{2}$ and clear denominators appropriately. This identity is probably quite classical but I don't know if it has a name; it's related to the Norlund-Rice integral. After substituting $z \mapsto - \frac{1}{z}$ and clearing denominators appropriately, it can be thought of in terms of one of the generating functions of the Stirling numbers of the second kind.