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 \[ \begin{align} f(x) = (1+x)^{-3} &\implies f(0) = 1 \\ f'(x) = -3(1+x)^{-4} &\implies f'(0) = -3 \\ f''(x) = (-3 \cdot -4)(1+x)^{-5} &\implies f''(0) = -3\cdot -4 \\ & \vdots \\ f^{(k)}(x) = -3 \cdot -4 \cdots (-3-k+1)(1+x)^{-3-k} &\implies f^{(k)}(0) = -3 \cdot -4 \cdots (-3-k+1). \end{align} \]

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

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