Negative Binomial Theorem

The binomial theorem for positive integer exponents $n$ can be generalized to negative integer exponents. This gives rise to several familiar Maclaurin series with numerous applications in calculus and other areas of mathematics.

$f(x) = (1+x)^{-3}$ is not a polynomial. While positive powers of $1+x$ can be expanded into polynomials, e.g. $(1+x)^3 = 1+3x+3x^2+x^3$, $f(x)$ cannot be, so there cannot be a finite sum of monomial terms that equals $f(x)$. But there is a way to recover the same type of expansion if infinite sums are allowed.

As a first approximation, since $f'(0)= -3$ by the power rule, the tangent line at $x = 0$ is $y = 1 -3x$. So for small $x$, $\frac1{(1+x)^3} \approx 1-3x.$ This approximation is already quite useful, but it is possible to approximate the function more carefully using series.

Expand $\frac1{(1+x)^3}$ as a Maclaurin series.

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The Maclaurin series for $f(x)$, wherever it converges, can be expressed as

$f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \cdots + \frac{f^{(k)}(x)}{k!} x^k + \cdots.$

Let $f(x) = \frac1{(1+x)^3}$. Applying the power rule repeatedly, we have $\begin{aligned} f(x) = (1+x)^{-3} &\implies f(0) = 1 \\ f'(x) = -3(1+x)^{-4} &\implies f'(0) = -3 \\ f''(x) = (-3 \cdot -4)(1+x)^{-5} &\implies f''(0) = -3\cdot -4 \\ & \vdots \\ f^{(k)}(x) = -3 \cdot -4 \cdots (-3-k+1)(1+x)^{-3-k} &\implies f^{(k)}(0) = -3 \cdot -4 \cdots (-3-k+1). \end{aligned}$

So the Maclaurin series becomes

$f(x) = 1 - 3x + \frac{-3 \cdot -4}{2!} x^2 + \cdots + \frac{-3 \cdot -4 \cdots (-3-k+1)}{k!} x^k + \cdots.$

This converges for $|x|<1$ by the ratio test. $_\square$

The above example generalizes immediately for all negative integer exponents $\alpha$. Let $\alpha$ be a real number and $k$ a positive integer. Define

$\binom{\alpha}{k} = \frac{\alpha(\alpha-1)\ldots(\alpha-k+1)}{k!} = \frac{\alpha !}{k!(\alpha-k)!},$

then the same analysis as in the example gives

Let $n$ be a positive integer. Then $\frac1{(1+x)^n} = 1 - nx + \frac{(-n)(-n-1)}2 x^2 + \cdots = \sum_{k=0}^{\infty} \binom{n+k-1}{k} (-1)^kx^k$ for $|x|<1$.

For $n = 2$, we have

$\binom{-2}{k} = \frac{(-2)(-3)\ldots(-k-1)}{k!} = (-1)^k(k+1).$

So

$\frac1{(1+x)^2} = \sum_{k=0}^{\infty} \binom{-2}{k} x^k = \sum_{k=0}^\infty (k+1)(-x)^k.$

Plugging in $x = -\frac12$, for instance, gives

$4 = \sum_{k=0}^{\infty} \frac{k+1}{2^k}.$