2
$\begingroup$

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.

|cite|improve this question
$\endgroup$
3
$\begingroup$

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.