How to calculate $\sum_{k=0}^{n} \binom{a+k}{a} x^k$

Question

As the title mentioned, I want to calculate \begin{equation} \sum_{k=0}^{n} \binom{a+k}{a} x^k, \end{equation} where $n$ and $a$ are both positive integer, and $x\in[0,1]$.

There is an obvious upper bound as \begin{equation} \sum_{k=0}^{n} \binom{a+k}{a} x^k<\sum_{k=0}^{\infty} \binom{a+k}{a} x^k=(1-x)^{-a-1}, \end{equation} which is too loose for my application. Therefore, is it possible to obtain the exact result of this sum? Or at least a strict upper bound tighter than the above one. BTW, I came to this sum from hypergeometric function (or imcomplete beta function with transform in form), so I do not want to go back.

Answer 1

(Long comment) I am skeptical about the existence of a simple closed form for the sum. However, we have some alternative representation of the series that might be possibly useful in the case

• $a$ if fixed and $n$ is increasing, or
• $x$ is close to $1$.

Below is such a representation:

\begin{align*} \sum_{k=0}^{n} \binom{a+k}{a} x^k &= \sum_{k=0}^{n} \binom{n+1+a}{n-k} x^{n-k}(1 - x)^{k} \\ &= \frac{1}{(1-x)^{a+1}} \left[ 1 - \sum_{k=0}^{a} \binom{n+1+a}{k} x^{n+1+a-k}(1 - x)^k \right] \end{align*}

In the last expression, tossing the entire inner sum away recovers the crude bound $\frac{1}{(1-x)^{a+1}}$, whereas keeping (part of) it gives a better upper bound.

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Also, for $x \in [0, 1]$ this representation admits the following reformulation in terms of probability: Let $X \sim \text{Binomial}(n+1+a, x)$ be a binomial random variable with parameterx $n+1+a$ and $x$. Then

\begin{align*} \sum_{k=0}^{n} \binom{a+k}{a} x^k &= \sum_{k=0}^{n} \binom{n+1+a}{n-k} x^{n-k}(1 - x)^{k} \\ &= \frac{1}{(1-x)^{a+1}} \left[ 1 - \mathbf{P}(X > n) \right]. \end{align*}

Hence, if $x$ is sufficiently close to $1$ so that $\mathbf{E}[X] > n$, we can invoke several tail estimates from the probability theory to give an upper bound.

Example. If $\mathbf{E}[X] > n$, or equivalently $x > \frac{n}{n+a+1}$, then by Hoeffding's inequality,

\begin{align*} 1 - \mathbf{P}(X > n) &= \mathbf{P}(X - \mathbf{E}[X] \leq -(\mathbf{E}[X] - n) ) \\ &\leq \exp\left( -\frac{2(\mathbf{E}[X] - n)^2}{n+a+1} \right) \\ &= \exp\left( -2(n+a+1)\left(x- \frac{n}{n+a+1}\right)^2 \right). \end{align*}

This is particularly useful when $a \gg n$.

Answer 2

Let $$f(x;a,n) = \sum_{k=0}^n \binom{a+k}{a} x^k. \tag{1}$$ Then $$f(x;a,\infty) = (1-x)^{-(a+1)}, \quad |x| < 1 \tag{2}$$ as this is a negative binomial series.

The evaluation of this sum for finite $n$ can be expressed in terms of the incomplete beta function: $$f(x;a,n) = \frac{1 - (n+1) {\operatorname{B}}_x(n+1, a+1) \binom{a+n+1}{a}}{(1-x)^{a+1}}, \quad |x| < 1, \tag{3}$$ where $${\operatorname{B}}_z(a,b) = \int_{t=0}^z t^{a-1} (1-t)^{b-1} \, dt \tag{4}$$ is the incomplete beta function. However, this is not a particularly helpful result, since the evaluation of $(4)$ and subsequently $(3)$ for positive integer values $a, n$ only leads back to $(1)$.

Answer 3

Here is my solution.

Let $f(a) = \sum_{k =0}^{n}{a+k \choose a}x^k$

$xf(a) = \sum_{k =0}^{n}{a+k \choose a}x^{k+1}$

$(1-x)f(a)= \sum_{k =0}^{n}{a+k \choose a}x^k - \sum_{k =0}^{n}{a+k \choose a}x^{k+1}$

$= 1 + \sum_{k =1}^{n}{a+k \choose a}x^k - \sum_{k =0}^{n}{a+k \choose a}x^{k+1}$ (Note k starts from 1 in the first summation.)

$= 1 + \sum_{k =1}^{n}{a+k \choose a}x^k - \sum_{k =1}^{n+1}{a+k-1 \choose a}x^{k}$ (Note some manipulations the second summation .)

$= 1 + \sum_{k =1}^{n}{a+k \choose a}x^k - \sum_{k =1}^{n}{a+k-1 \choose a}x^{k}-{{a+n}\choose a} x^{n+1}$

$= 1 + \sum_{k =1}^{n}\left({a+k \choose a}-{a+k-1 \choose a}\right)x^k -{{a+n}\choose a} x^{n+1}$

$=1 + \sum_{k =1}^{n}{a+k-1 \choose a-1}x^k -{{a+n}\choose a} x^{n+1}$ (because $\left({a+k \choose a}-{a+k-1 \choose a}\right) = {a+k-1 \choose a-1}$)

$=\sum_{k =0}^{n}{a+k-1 \choose a-1}x^k -{{a+n}\choose a} x^{n+1}$

$= f(a-1) - {{a+n}\choose a} x^{n+1}$

Then we have

$f(a) = \frac{1}{1-x} f(a-1) - {{a+n}\choose a} \frac{x^{n+1}}{1-x}$

$= \frac{1}{1-x} f(a-1) - {{a+n}\choose n} \frac{x^{n+1}}{1-x}$

Expand this,

$f(a) = \frac{1}{1-x} \left(\frac{1}{1-x} f(a-2) - {{a-1+n}\choose n} \frac{x^{n+1}}{1-x}\right) - {{a+n}\choose n} \frac{x^{n+1}}{1-x}$

$=\frac{1}{(1-x)^2}f(a-2)- {{a-1+n}\choose n}\frac{x^{n+1}}{(1-x)^2}- {{a+n}\choose n} \frac{x^{n+1}}{1-x} $

$=\ ...$

$=\frac{1}{(1-x)^a}f(0)- {{1+n}\choose n} \frac{x^{n+1}}{(1-x)^a} - {{2+n}\choose n} \frac{x^{n+1}}{(1-x)^{a-1}} - ... - {{a+n}\choose n} \frac{x^{n+1}}{1-x}$

$= \frac{1}{(1-x)^a}f(0) - x^{n+1}\sum_{k = 1}^{a}{{a+n+1-k} \choose n} \frac{1}{(1-x)^k}$

Now we have $f(0) = \sum_{k=0}^{n}x^k= \frac{1-x^{n+1}}{1-x}$,

$\sum_{k =0}^{n}{a+k \choose a}x^k=f(a) = \frac{1-x^{n+1}}{(1-x)^{a+1}} - x^{n+1}\sum_{k = 1}^{a}{{a+n+1-k} \choose n} \frac{1}{(1-x)^k}$

I've done simple verifications as follows.

Let $a = 2, n = 2$,

$f(2) = 1 + 3x + 6x^2$ (by plugging in the original form.)

$f(2) = \frac{1-x^{3}}{{(1-x)}^3}- x^{3}\sum_{k = 1}^{2}{{5-k} \choose 2} \frac{1}{(1-x)^k}$ (by plugging in the derived result.)

$= \frac{1+x+x^2}{(1-x)^2}- x^3\left(\frac{6}{1-x} + \frac{3}{(1-x)^2} \right)$

$=\frac{6x^4 - 9x^3+x^2+x+1}{(1-x)^2}$

$=1 + 3x + 6x^2$

An obvious bound:

For $0<x<1, \sum_{k =0}^{n}{a+k \choose a}x^k <\frac{1-x^{n+1}}{(1-x)^{a+1}}$.

Another bound:

For $0<x<1, \sum_{k =0}^{n}{a+k \choose a}x^k <\frac{1-x^{n+1}}{(1-x)^{a+1}}-x^{n+1}\sum_{k = 1}^{a}\frac{1}{(1-x)^k}= \frac{1-x^{n+1}}{(1-x)^{a+1}} - x^n \left((1-x)^{-a}-1\right)$ (by sum of geometric sequence.)

Finally given that ${a+n+1-k} \choose n$ appears in the result, we can derive more interesting bounds based on results of the bounds of general ${n \choose k}$. Some discussions are found here