# Calculation involving binomial coefficients

I could use somebody's help to understand a calculation:

$$\begin{array}{lcl} \sum\limits_{s=0}^{\infty} z^s\int_0^1 \frac{t(t+1)..(t+s-1)}{s!}dt &=& \sum\limits_{s=0}^{\infty} z^s \int_0^1 {{t+s-1}\choose{s}}dt \\ &=& \sum\limits_{s=0}^{\infty} z^s(-1)^s \int_0^1 {{-t}\choose{s}}dt\\ &=& \int_0^1\sum\limits_{s=0}^{\infty} {{-t}\choose{s}}(-z)^sdt\\ &=& \int_0^1(1-z)^{-t}dt \end{array}$$

I can take the first equality to be a definition, even if I knew this kind of expression for the binomial coefficients only for integers ${n}\choose{k}$.The second equality is a bit mysterious to me. The inversion of the summation and the integral I would have done it from the beginning provided $z \geq 0$ by monotone convergence theorem. The last step is also not clear to me.

Any help will be appreciated, thank you!

Let's go step by step and see what the calculation is.

1.First is simply using the definition of $\binom{n}{r}$

1. The second one used $\int_a^b f(x)dx= \int_a^b f(a+b-x) dx$, which gives us $\binom{s-t}{s}$, which clearly is $(-1)^s \binom{-t}{s}$

2. The third step simply involves taking the entire thing together, which brings the summation inside the integral.

3. The last uses the binomial theorem for negative powers$(1+x)^{-r}=1+rx+ \frac{r(r-1)x}{2}...$ or simply $\sum_{r=1}^{\infty}\binom{-n}{r} (x)^r$

1. The interchange of infinite summation and integration is possible if $|z|<1$, this is called uniform continuity.
2. If you have an expression of the form $\binom{-k}{s}$, just set $-k=\alpha$ and use the Binomial expansion (As it's done in your first step). You get $(-1)^s \binom{k+s}{s}$.
3. This is called Generalized Binomial expansion: $(1+a)^{-r} = \sum_{k=0}^{\infty} \binom{-r}{k} a^k$. This is valid for $|a|<1$.