Evaluating $\sum_{i=0}^{H-1} \dfrac{(-1)^i}{i! (H-1-i)! (k+i)}$ I've been struggling to directly find a closed form to the following sum
$$
\sum_{i=0}^{H-1} \dfrac{(-1)^i}{i! (H-1-i)! (K+i)}\,,\qquad K,H\in\mathbb{N}-\{0\}\,.
$$
I've found the result indirectly by evaluating the integral:
$$
\int_0^y x^{K-1}(y-x)^{H-1}\,dx\,,
$$
By expanding the binomial via Newton formula in fact you can find:
$$
\int_0^y x^{K-1}(y-x)^{H-1}\,dx=y^{H+K-1}(H-1)!\sum_{i=0}^{H-1} \dfrac{(-1)^i}{i! (H-1-i)! (K+i)}\,,
$$
knowing that the result of the integral is $\dfrac{(K-1)!(H-1)!}{(K+H-1)!}y^{K+H-1}$, you can then find the result:
$$
\sum_{i=0}^{H-1} \dfrac{(-1)^i}{i! (H-1-i)! (k+i)}=\dfrac{(K-1)!}{(H+K-1)!}=\dfrac{\Gamma(K)}{\Gamma(H+K)}=\dfrac{B(K,H)}{\Gamma(H)}
$$
Is there any other more "direct" way to evaluate that sum?
 A: This is really just a rehash of your method, but I would approach this as follows. Let
$$f(x)= \sum_{i=0}^{H-1} \dfrac{(-1)^i x^{K+i}}{i! (H-1-i)! (K+i)} $$
Then
$$f'(x)= \sum_{i=0}^{H-1} \dfrac{(-1)^i x^{K+i-1}}{i! (H-1-i)! }$$
$$= \frac{x^{K-1}}{(H-1)!}\sum_{i=0}^{H-1} \dfrac{ (H-1)! }{i! (H-1-i)! } (-x)^i $$
$$ = \frac{x^{K-1}}{(H-1)!}(1-x)^{H-1} $$
and integrating from 0 to 1:
$$f(1)-f(0)= \frac{1}{(H-1)!} B(K,H)=\frac{1}{(H-1)!}\dfrac{\Gamma(K)\Gamma(H)}{\Gamma(K+H)}= \dfrac{(K-1)!}{(K+H-1)!}$$
Where $B$ is the Beta function, $f(0)=0$ and $f(1)$ is our required sum.
A: Using the binomial theorem,
$$\int_0^1 (1-(1-\alpha)x)^{K+H-2}dx=\int_0^1 (1-x+\alpha x)^{K+H-2}dx\\ =\sum_{n=0}^{K+H-2}\binom{K+H-2}{n}\alpha^n\int_0^1  x^n (1-x)^{K+H-n-2}dx$$
Then we calculate
$$\int_0^1(1-(1-\alpha)x)^{K+H-2}dx =\frac{1}{K+H-1}\frac{1-\alpha^{K+H-1}}{1-\alpha}=\frac{1}{K+H-1}\sum_{n=0}^{K+H-2}\alpha^n$$
Comparing coefficients of $\alpha^{H-1}$, we get
$$\frac{1}{H+K-1}=\binom{K+H-2}{H-1}\int_0^1  x^n (1-x)^{K+H-n-2}dx \\ =\binom{K+H-2}{H-1}(H-1)!\sum_{n=0}^{H-1}\frac{(-1)^{n}}{n!(H-n-1)!(K+n)}$$
So, $$\sum_{n=0}^{H-1}\frac{(-1)^{n}}{n!(H-n-1)!(K +n)}=\frac{(K-1)!}{(H+K-1)!}$$
