prove: $\sum_{r=0}^{ ∞}\frac{{{r+k}\choose{k}}}{(r+k)(r+k-1)}x^r=\frac{\left(k- 2\right)!}{k!}\cdot\frac{1}{\left(1-x\right)^{\left(k-1\right)}}$

Question

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

enter image description here

Calvin Lin · Answer 1 · 2019-12-22T15:53:05.967

1

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

$$ \begin{array}{l l l } &=& \frac{(k-2)!}{k!}\sum \frac{(r+k-2)!}{r!(k-2)!} x^r \\ &= &\sum \frac{(r+k-2)!}{r!k!} x^r \\ &=& \sum \frac{1}{(r+k)(r+k-1)} \times \frac{(r+k)!}{r!k!} x^r \\ &=& \sum \frac{1}{(r+k)(r+k-1)} {r+k \choose k } x^r \\ &=& LHS \end{array} $$

edited Dec 22 '19 at 15:53

answered Dec 18 '19 at 15:55

Calvin Lin

68,864

yes you are right, but I still cannot get the finall result (the pic is not my work) – Absurd Dec 18 '19 at 16:06
@Absurd All that was left was direct algebraic manipulation, which I have filled in. To me, the crux was in expanding the RHS as a power series via the (negative) binomial theorem. – Calvin Lin Dec 22 '19 at 15:54
but it's not like the finall answer – Absurd Dec 22 '19 at 17:29
What is not like the final answer? Please elaborate. – Calvin Lin Dec 22 '19 at 23:54

prove: $\sum_{r=0}^{ ∞}\frac{{{r+k}\choose{k}}}{(r+k)(r+k-1)}x^r=\frac{\left(k- 2\right)!}{k!}\cdot\frac{1}{\left(1-x\right)^{\left(k-1\right)}}$

1 Answers1