Showing an identity via induction I'm conjecturing that
$$\sum_{j=0}^k \binom{k}{j}\frac{(-1)^j}{j\theta+1}=\frac{k!\theta^k}{\prod_{i=1}^k(i\theta+1)} \quad (1)$$
where $\theta>0$ and $k\in\mathbb N^*$.
I checked that it works for $k\in\{1,2,3,4\}$. But I'm struggling to show it works for $k+1$ from the inductive hypothesis $(1)$. I used the Pascal's rule, but didn't succeed.
Does $(1)$ holds? How can we prove it?
 A: Firstly, $\frac{1}{j\theta+1}=\int_0^1x^{j\theta}dx$ rewrites the LHS as $\int_0^1(1-x^\theta)^kdx$ by the binomial theorem. If you're familiar with Beta and Gamma functions, this is $\tfrac{1}{\theta}\operatorname{B}(\tfrac{1}{\theta},\,k+1)=\frac{k!\Gamma(1+\tfrac{1}{\theta})}{\Gamma(k+1+\tfrac{1}{\theta})}$. Now use $\Gamma(s+1)=s\Gamma(s)$ to prove by induction this is equal to your RHS.
A: In trying to evaluate
$$S_k(\theta) = \sum_{j=0}^k {k\choose j} \frac{(-1)^j}{j\theta+1}$$
where $\theta\gt 0$ we introduce
$$f(z) = \frac{k!\times (-1)^k}{z\theta+1}
\prod_{q=0}^k \frac{1}{z-q}.$$
This has poles at $0,1,\ldots k$ and $-1/\theta.$ With $\theta\gt 0$
there is no overlap with the poles at the integers because those are
nonnegative and $-1/\theta \lt 0.$
Observe that with $0\le j\le k$
$$\mathrm{Res}_{z=j} f(z)
= \frac{k!\times (-1)^k}{j\theta+1}
\prod_{q=0}^{j-1} \frac{1}{j-q} \prod_{q=j+1}^k \frac{1}{j-q}
\\ = \frac{k!\times (-1)^k}{j\theta+1}
\frac{1}{j!} \frac{(-1)^{k-j}}{(k-j)!}
= {k\choose j} \frac{(-1)^j}{j\theta+1}.$$
It follows that
$$S_k(\theta) = \sum_{j=0}^k \mathrm{Res}_{z=j} f(z).$$
Now residues sum to zero and the residue at infinity is zero by
inspection. We get
$$S_k(\theta) = - \mathrm{Res}_{z=-1/\theta} f(z)
= - \frac{1}{\theta} \mathrm{Res}_{z=-1/\theta}
\frac{k!\times (-1)^k}{z+1/\theta}
\prod_{q=0}^k \frac{1}{z-q}
\\ = - \frac{1}{\theta} k! (-1)^k
\prod_{q=0}^k \frac{1}{-1/\theta - q}
= \frac{1}{\theta} k! 
\prod_{q=0}^k \frac{1}{1/\theta + q}
\\ = \frac{k!\times \theta^k}{\prod_{q=0}^k (1+q\theta)}.$$
This is the claim.
