Closed form for $\sum_{n=0}^{\infty} \binom{n+k}{k} x^n$ (Negative Binomial Theorem)

Question

I was wondering if there is also a closed expression for the series

$$\sum_{n=0}^{\infty} \binom{n+k}{k} x^n$$ where $|x|<1.$

A few examples suggest that the answer is $\frac{1}{(1-x)^{k+1}}$ but I don't see how to show this.

Answer 1

You can also proceed by adding a variable. Let:

$$f(x,y) := \sum_{k=0}^{+ \infty} y^k \sum_{n=0}^{+ \infty} \binom{n+k}{k} x^n.$$

Then:

$$f(x,y) = \sum_{k=0}^{+ \infty} \sum_{n=k}^{+ \infty} \binom{n}{k} \left( \frac{y}{x} \right)^k x^n = \sum_{n=0}^{+ \infty} x^n \sum_{k=0}^n \binom{n}{k} \left( \frac{y}{x} \right)^k = \sum_{n=0}^{+ \infty} x^n \left( 1+\frac{y}{x} \right)^n = 1/(1-y-x).$$

If you take the $k$-th partial derivative with respect to $y$ in $(x,0)$ and divide by $k!$, you get:

$$\sum_{n=0}^{+ \infty} \binom{n+k}{k} x^n = \frac{1}{(1-x)^{k+1}}.$$

Now, you just have to be careful about the radius of convergence of such series. Taking $|y| < 1-|x|$ should do the trick.

Answer 2

Negative Binomial Coefficients

As shown in this answer, $$ \binom{-n}{k}=(-1)^k\binom{n+k-1}{k} $$ Substituting $n\mapsto n+1$ then multiplying by $(-1)^k$ gives $$ (-1)^k\binom{-n-1}{k}=\binom{n+k}{k} $$ Thus, $$ \begin{align} \sum_{k=0}^\infty\binom{n+k}{k}x^k &=\sum_{k=0}^\infty(-1)^k\binom{-n-1}{k}x^k\\ &=\sum_{k=0}^\infty\binom{-n-1}{k}(-x)^k\\[4pt] &=(1-x)^{-n-1}\\[7pt] &=\frac1{(1-x)^{n+1}} \end{align} $$

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Differentiating A Geometric Series

Start with $$ \frac1{1-x}=\sum_{k=0}^\infty x^k $$ and take $n$ derivatives $$ \begin{align} \frac{n!}{(1-x)^{n+1}} &=\sum_{k=0}^\infty k(k-1)\cdots(k-n+1)x^{k-n}\\ &=\sum_{k=n}^\infty k(k-1)\cdots(k-n+1)x^{k-n}\\ &=\sum_{k=0}^\infty(k+n)(k+n-1)\cdots(k+1)x^k\\ &=\sum_{k=0}^\infty\frac{(k+n)!}{k!}x^k \end{align} $$ Finally, divide by $n!$ to get $$ \frac1{(1-x)^{n+1}}=\sum_{k=0}^\infty\binom{k+n}{k}x^k $$

Answer 3

One has: $$\frac{1}{(1-x)^{k+1}}=\prod_{i=1}^{k+1}\frac{1}{1-x}=\prod_{i=1}^{k+1}\sum_{n=0}^{+\infty}x^n.$$ Now, let define: $$\forall n\in\mathbb{N},a_n:=\#\left\{(k_i)_{i\in\{1,\ldots,k+1\}}\in\mathbb{N}^{k+1}\textrm{ s.t. }\sum_{i=1}^{k+1}k_i=n.\right\}.$$ Hence, one has: $$\frac{1}{(1-x)^{k+1}}=\sum_{n=0}^{+\infty}a_nx^n.$$ Your problem is reduced to a combinatorial problem: for every $n\in\mathbb{N}$, you need to find $a_n$, you will find: $$a_n={n+k\choose k}.$$

Answer 4

A few examples suggest that the answer is $~\dfrac{1}{(1-x)^{k+1}}~,~$ but I don't see how to show this.

Of course you don't ! And do you know why ? Because of the n in $\displaystyle{\color{red}n+k\choose k}.~$ That cute little

culprit throws a wrench into all our efforts. After all, even if we were to expand $~\dfrac1{(1-x)^{k+1}}~$

into its binomial series, we'd still get a constant term $($in this case, $-k-1)$ there. Instead, we

have $n+k,~$ where n is variable. Unless we could show that $\displaystyle{n+k\choose k}=(-1)^n{-k-1\choose n}.~$

To do that, just rewrite $~\displaystyle{-k-1\choose n}~$ explicitly as

$$\dfrac{\Big[(-k-1)-0\Big]\Big[(-k-1)-1\Big]\cdots\Big[(-k-1)-(n-1)\Big]}{1\cdot2\cdot3\cdots n},$$

and factor $-1$ out of each term in the numerator, then compare the expression with that of

$${n+k\choose k}~=~\dfrac{\Big[(n+k)-0\Big]\Big[(n+k)-1\Big]\cdots\Big[(n+k)-(k-1)\Big]}{1\cdot2\cdot3\cdots k}~.$$

Depending on which of the two numbers k and n is greater, one of the two expressions will
suffer cancellations in both its numerator and denominator.