I'm trying to simplify the following formula, but I keep failing to do.
$$ \sum_{k=0}^{\infty} \binom{k + r - 1}{k} x^k $$
I tried:
From the Hockey-stick identity,
$$ \sum_{k=0}^{n} \binom{k + r - 1}{k} = \binom {n+r}{n}$$
But I have no clue to apply the fact to simplify the above formula.
Is there anyone to give a hint for it?
Write it as $\sum_{k=0}^\infty \binom{k+r-1}{r-1}x^k$. Then Newton's Generalized Binomial Theorem gives:
$$\frac{1}{(1-x)^r}=\sum_{k=0}^\infty \binom{k+r-1}{r-1}x^k.$$
If you need to prove this identity, simply differentiate the left-hand-side $k$ times and evaluate at 0, for each coefficient on the r.h.s.
Using Binomial Coefficients for Negative Exponents and Newton's Generalized Binomial Theorem: $$ \begin{align} \sum_{k=0}^\infty\binom{k+r-1}{k}x^k &=\sum_{k=0}^\infty(-1)^k\binom{-r}{k}x^k\\[6pt] &=(1-x)^{-r} \end{align} $$
Notice that if $D$ is the complex unit open disk and $r\in\mathbb{N}$, then the function $$f:D\rightarrow\mathbb{C}, z\mapsto\frac{1}{(1-z)^r}$$ is well defined and holomorphic. So the McLaurin series of $f$ converges uniformly on the compact subsets of the open disk to $f$. In particular, the McLaurin series of $f$ converges pointwise in the whole open disk to $f$. Then $$\forall z \in D, f(z)=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}z^k.$$ However $$\forall k\in\mathbb{N}_0, \frac{f^{(k)}(0)}{k!}=\frac{\prod_{j=0}^{k-1}(r+j)}{k!}=\frac{(k+r-1)!}{k!(r-1)!}=\binom{k + r - 1}{k},$$ where we used the convention that $\prod_{j=0}^{-1}(r+j)=1$ and so: $$\forall z\in D, \frac{1}{(1-z)^r}=f(z)=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}z^k= \sum _ {k=0}^{\infty}\binom{k + r - 1}{k}z^k.$$
$$ \binom{k+r-1}{k} = \frac{(k-1+r)(k-2+r)\dotsm (1+r)(r)}{k!} = (-1)^k\frac{(-r-k+1)(-r-k-2)\dotsm(-r-1)(-r)}{k!} = (-1)^k \binom{-r}{k}, $$ so the generalised binomial theorem gives $$ \sum_{k=0}^{\infty} \binom{k+r-1}{k} x^k = (1+x)^{-r}. $$
