Sum of diagonal binomial coefficients polynomial

I'm having trouble evaluating the following sum, for constant $$n$$ and $$-1 \leq x \leq 1$$:

$$\sum\limits_{k = 0}^{\infty} {n+k \choose k}x^k$$

I know the series must converge by the ratio test, since:

$$\lim\limits_{k \to \infty}\left|\frac{{n+k+1 \choose k+1}x^{k+1}}{{n+k \choose k}x^k}\right| = \lim\limits_{k \to \infty}\frac{n+k+1}{k+1}\left|x\right| = \left|x\right| < 1.$$

However, I'm not sure where to go from here.

Note that \begin{align} \binom{n+k}{k}&=\frac{(n+k)(n+k-1)\dots (n+1)}{k!}\\ &=(-1)^{k}\frac{(-n-1)(-n-2)\dots(-n-k)}{k!}\\ &=(-1)^k\binom{-n-1}{k} \end{align} so the series is the binomial expansion of $$(1-x)^{-n-1}$$.
• If $n$ is a nonnegative integer, then for $k>n$ the numerator of $\binom{n}{k}$ would contain a factor $0$ (the factor $n-n$) whereas the denominator $k!$ is nonzero. Hence the infinite series $\sum_k\binom{n}{k}x^k$ has only finitely many nonzero terms. – user10354138 Dec 6 '18 at 6:56