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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.

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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 $(1-x)^{-n-1}$.

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  • $\begingroup$ Thanks! I've never seen this trick before but that is so powerful... I thought the answer would end up being through a complicated generating function. Does it matter that in the 1st series k goes to infinity but in the second it is effectively bounded? Or are binomial coefficients consistent even when k exceeds n? $\endgroup$ – Aaron Z. Dec 6 '18 at 6:52
  • $\begingroup$ 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. $\endgroup$ – user10354138 Dec 6 '18 at 6:56
  • $\begingroup$ I see. Thanks again for the beautiful solution. $\endgroup$ – Aaron Z. Dec 6 '18 at 7:03

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