Evaluate $ 1 + \sum_{i=0}^{n-2} [ \prod_{k=0}^{i} (n-k)/(r-k) ] $ Is there a way in which the expression
$$ 1 + \sum_{i=0}^{n-2}  \prod_{k=0}^{i} \dfrac{n-k}{r-k}  $$
may be simplified?
It would simplify some equations I am working with.
 A: First, using recurrence equation for Euler's $\Gamma$-function:
$$
   \prod_{k=0}^m \frac{n-k}{r-k} = \frac{\Gamma(n+1)}{\Gamma(n-m)} \cdot \frac{\Gamma(r-m)}{\Gamma(r+1)}
$$
Then, changing summation variable as $m \to n-2-m$:
$$
   1 + \sum_{m=0}^{n-2} \frac{\Gamma(n+1)}{\Gamma(n-m)} \cdot \frac{\Gamma(r-m)}{\Gamma(r+1)} = 1 + \sum_{m=0}^{n-2} \frac{\Gamma(n+1)}{\Gamma(n-(n-2-m))} \cdot \frac{\Gamma(r-(n-m-2))}{\Gamma(r+1)} = \\
1 + \frac{\Gamma(n+1)}{\Gamma(r+1)}  \sum_{m=0}^{n-2} \frac{\Gamma(m+r+2-n)}{\Gamma(m+2)} 
 = 1 + \frac{\Gamma(n+1)}{\Gamma(r+1)}  \sum_{m=0}^{n-2} \frac{\Gamma(m+r+2-n)}{\Gamma(m+2)} 
$$
Now the last sum is just a hypergeometric series:
$$
   \frac{\Gamma(n+1)}{\Gamma(r+1)}  \sum_{m=0}^{n-2} \frac{\Gamma(m+r+2-n)}{\Gamma(m+2)}  = 
   \frac{\Gamma(n+1) \Gamma(1+r-n)}{\Gamma(r+1)}  \sum_{m=0}^{n-2} \frac{\Gamma(m+r+2-n)}{\Gamma(m+2) \Gamma(1+r-n)}  = \\
    \frac{1}{\binom{r}{n}} \sum_{m=0}^{n-2} \binom{m+r-n+1}{m+1} = \frac{n}{r-n+1} - \frac{1}{\binom{r}{n}}
$$
The last equality follows because $s_{m+1} - s_m = \binom{m+r-n+1}{m+1}$, where
$$
    s_m = \frac{m+1}{r-n+1}\binom{m+r-n+1}{m+1}
$$
And therefore the original sum telescopes 
$$ \sum_{m=0}^{n-2} \binom{m+r-n+1}{m+1} = \sum_{m=0}^{n-2} (s_{m+1}-s_m)  = s_{n-1} - s_0
$$
It now remains to compute $s_{m+1}-s_m$:
$$
  s_{m+1} = \frac{m+2}{r-n+1} \binom{m+r-n+2}{m+2} = \frac{m+2}{r-n+1} \cdot \frac{m+r-n+2}{m+2} \binom{m+r-n+1}{m+1} = \\ s_m + \binom{m+r-n+1}{m+1}
$$
A: I believe your sum comes out to:
$$ \frac{r+1}{r-n+1} - \frac{1}{\binom{r}{n}}$$
We can prove this using Beta integrals. We have the identity:
$$\frac{1}{\binom{r}{k}} = (r+1)\int_{0}^{1} t^{r-k} (1-t)^k dt$$
What you have is
$$1+\sum_{k=1}^{n-1} \frac{\binom{n}{k}}{\binom{r}{k}}$$
Now
$$\sum_{k=1}^{n-1} \frac{\binom{n}{k}}{\binom{r}{k}} = \sum_{k=1}^{n-1} (r+1)\int_{0}^{1} \binom{n}{k} t^{r-k} (1-t)^k dt = $$
$$ (r+1)\int_{0}^{1} \sum_{k=1}^{n-1} \binom{n}{k} t^{r-k} (1-t)^k dt  = $$
$$ (r+1)\int_{0}^{1} t^{r-n} \sum_{k=1}^{n-1} \binom{n}{k} t^{n-k} (1-t)^k dt  = $$
$$ (r+1)\int_{0}^{1} t^{r-n} (1 - t^n - (1-t)^n) dt  = $$
$$ (r+1)\int_{0}^{1} t^{r-n} - t^r - t^{r-n}(1-t)^n dt =  $$
$$ \frac{r+1}{r-n+1} - 1 - (r+1)\int_{0}^{1}t^{r-n}(1-t)^n dt =$$
$$ \frac{r+1}{r-n+1} - 1 - \frac{1}{\binom{r}{n}}$$
(at the last step we used the Beta integral identity mentioned earlier).
Thus the sum you seek is
$$ \frac{r+1}{r-n+1} - \frac{1}{\binom{r}{n}}$$
Here are a couple of answers which use Beta integral:
Formula for the harmonic series $H_n = \sum_{k=1}^n 1/k$ due to Gregorio Fontana
https://math.stackexchange.com/a/6533/1102
