prove: $$\sum_{r=0}^{ ∞}\frac{{{r+k}\choose{k}}}{(r+k)(r+k-1)}x^r=\frac{\left(k- 2\right)!}{k!}\cdot\frac{1}{\left(1-x\right)^{\left(k-1\right)}}$$ I tried to expand the sum such that: $$\sum_{r=0}^{ ∞}\frac{{{r+k}\choose{k}}}{(r+k)(r+k-1)}x^r=\frac{1}{k!}\sum_{r=0}^{ ∞}\frac{(r+k- 2)!}{r!}x^r$$ I think this sum can be seen as a geometric sum, but I cannot reach that.
here is a photo which maybe helpful
You're nearly there!
By the binomial theorem, the RHS can be expanded as
$$ \frac{(k-2)!}{k!}\sum {r+k-2 \choose k-2} x^r,$$
It's an equality if the coefficients of $x^r$ are the same. Are they?
Here's the working
$$ \begin{array}{l l l } &=& \frac{(k-2)!}{k!}\sum \frac{(r+k-2)!}{r!(k-2)!} x^r \\ &= &\sum \frac{(r+k-2)!}{r!k!} x^r \\ &=& \sum \frac{1}{(r+k)(r+k-1)} \times \frac{(r+k)!}{r!k!} x^r \\ &=& \sum \frac{1}{(r+k)(r+k-1)} {r+k \choose k } x^r \\ &=& LHS \end{array} $$