Binomial series

In mathematics, the binomial series is the Maclaurin series for the function $f$ given by $f(x)=(1+x)^{\alpha}$, where $\alpha \in \mathbb{C}$ is an arbitrary complex number. Explicitly,

$\begin{align} (1 + x)^\alpha &= \sum_{k=0}^{\infty} \; {\alpha \choose k} \; x^k \qquad\qquad\qquad (1) \\ &= 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!} x^2 + \cdots, \end{align}$

and the binomial series is the power series on the right hand side of (1), expressed in terms of the (generalized) binomial coefficients

${\alpha \choose k} := \frac{\alpha (\alpha-1) (\alpha-2) \cdots (\alpha-k+1)}{k!}.$

Special cases

If α is a nonnegative integer n, then the (n + 2)th term and all later terms in the series are 0, since each contains a factor (n − n); thus in this case the series is finite and gives the algebraic binomial formula.

The following variant holds for arbitrary complex β, but is especially useful for handling negative integer exponents in (1):

$\frac{1}{(1-z)^{\beta+1}} = \sum_{k=0}^{\infty}{k+\beta \choose k}z^k.$

To prove it, substitute x = −z in (1) and apply a binomial coefficient identity, which is,

${-\beta-1 \choose k} = (-1)^k {k+\beta \choose k}$.

Convergence

Conditions for convergence

Whether (1) converges depends on the values of the complex numbers α and x. More precisely:

1. If | x | < 1, the series converges absolutely for any complex number α.
2. If x = −1, the series converges absolutely if and only if either Re(α) > 0 or α = 0.
3. If | x | = 1 and x ≠ −1, the series converges if and only if Re(α) > −1.
4. If | x | > 1, the series diverges, unless α is a non-negative integer (in which case the series is finite).

Assume now that $\alpha$ is not a non-negative integer and that $|x| = 1$. We make the following additional observations, which follow from the ones above:

• If Re(α) > 0, the series converges absolutely.
• If -1 < Re(α) ≤ 0, the series converges conditionally if x ≠ −1 and diverges if x = −1.
• If Re(α) ≤ -1, the series diverges.

Identities to be used in the proof

The following hold for any complex number α:

$\begin{align} {\alpha \choose 0} &= 1, \\ {\alpha \choose k+1} &= {\alpha\choose k}\,\frac{\alpha-k}{k+1}, &\qquad\qquad&(2) \\ {\alpha \choose k-1} + {\alpha\choose k} &= {\alpha+1 \choose k}. &&(3) \end{align}$

Unless α is a nonnegative integer (in which case the binomial coefficients vanish as k is larger than α), a useful asymptotic relationship for the binomial coefficients is, in Landau notation:

${\alpha \choose k} = \frac{(-1)^k} {\Gamma(-\alpha)k^ {1+\alpha} } \,(1+o(1)), \quad\text{as }k\to\infty. \qquad\qquad(4)$

This is essentially equivalent to the Gauss limit for the Gamma function:

$\Gamma(z) = \lim_{k \to \infty} \frac{k! \; k^z}{z \; (z+1)\cdots(z+k)}, \qquad$

and implies immediately the coarser bounds

$\frac {m} {k^{1+\operatorname{Re}\,\alpha}}\le \left|{\alpha \choose k}\right| \le \frac {M} {k^{1+\operatorname{Re}\,\alpha}}, \qquad\qquad(5)$

for some positive constants m and M independent of k, which are in fact sufficient for our needs. The simpler bounds (5) may also be obtained by means of elementary inequalities (see the addendum below for the latter inequality).

Proof

To prove (i) and (v), apply the ratio test and use formula (2) above to show that whenever α is not a nonnegative integer, the radius of convergence is exactly 1. Part (ii) follows from formula (5), by comparison with the p-series

$\sum_{k=1}^\infty \; \frac {1} {k^p}, \qquad$

with p = 1 + Re(α). To prove (iii), first use formula (3) to obtain

$(1 + x) \sum_{k=0}^n \; {\alpha \choose k} \; x^k =\sum_{k=0}^n \; {\alpha+1\choose k} \; x^k + {\alpha \choose n} \;x^{n+1},$

and then use (ii) and formula (5) again to prove convergence of the right-hand side when Re(α) > −1 is assumed. On the other hand, the series does not converge if |x| = 1 and Re(α) ≤ −1, because in that case, for all k,

$\left|{\alpha \choose k}\; x^k \right| \geq 1,$

completing the proof of (iii). Also, the identity above, for x=-1 and with α+1 in place of α writes

$\sum_{k=0}^n \; {\alpha\choose k} \; (-1)^k = {\alpha-1 \choose n} \;(-1)^n,$

whence (iv) follows using (5) again.

Summation of the binomial series

The usual argument to compute the sum of the binomial series goes as follows. Differentiating term-wise the binomial series within the convergence disk |x| < 1 and using formula (1), one has that the sum of the series is an analytic function solving the ordinary differential equation (1 + x)u'(x) = α u(x) with initial data u(0) = 1. The unique solution of this problem is the function u(x) = (1 + x)^α, which is therefore the sum of the binomial series, at least for |x| < 1. The equality extends to |x| = 1 whenever the series converges, as a consequence of Abel's theorem and by continuity of (1 + x)^α.

History

The first results concerning binomial series for other than positive-integer exponents were given by Sir Isaac Newton in the study of areas enclosed under certain curves. John Wallis built upon this work by considering expressions of the form y = (1 − x²)^m where m is a fraction. He found that (written in modern terms) the successive coefficients c_k of (-x²)^k are to be found by multiplying the preceding coefficient by $\tfrac{m-(k-1)}k$ (as in the case of integer exponents), thereby implicitly giving a formula for these coefficients. He explicitly writes the following instances

$(1-x^2)^{1/2}=1-\frac{x^2}2-\frac{x^4}8-\frac{x^6}{16}\cdots$

$(1-x^2)^{3/2}=1-\frac{3x^2}2+\frac{3x^4}8+\frac{x^6}{16}\cdots$

$(1-x^2)^{1/3}=1-\frac{x^2}3-\frac{x^4}9-\frac{5x^6}{81}\cdots$

The binomial series is therefore sometimes referred to as Newton's binomial theorem. Newton gives no proof and is not explicit about the nature of the series; most likely he verified instances treating the series as (again in modern terminology) formal power series. Later, Niels Henrik Abel discussed the subject in a memoir, treating notably questions of convergence.

Elementary bounds on the coefficients

In order to keep the whole discussion within elementary methods, one may derive the asymptotics (5) proving the inequality

$\left|{\alpha \choose k} \right|\leq\frac {M}{k^{1+\mathrm{Re}\,\alpha}},\qquad\forall k\geq1$

with a M independent of k as follows. By the inequality of arithmetic and geometric means

$\left|{\alpha \choose k} \right|^2=\prod_{j=1}^k \left|1-\frac{1+\alpha}{j}\right|^2 \leq \left( \frac{1}{k}\sum_{j=1}^{k} \left|1-\frac{1+\alpha}{j}\right|^2 \right)^k.$

Using the expansion

$\textstyle |1-\zeta|^2=1-2\mathrm{Re}\,\zeta +|\zeta|^2$

the latter arithmetic mean writes

$\frac{1}{k}\sum_{j=1}^{k} \left|1-\frac{1+\alpha}{j}\right|^2= 1+\frac{1}{k}\left(- 2(1+\mathrm{Re}\,\alpha) \sum_{j=1}^{k}\frac{1}{j}+|1+\alpha|^2\sum_{j=1}^{k}\frac{1}{j^2}\right)\ .$

To estimate its kth power we then use the inequality

$\left(1+\frac{r}{k}\right)^k\leq \mathrm{e}^r,$

that holds true for any real number r as soon as 1 + r/k ≥ 0. Moreover, we have elementary bounds for the sums:

$\log k\leq\sum_{j=1}^k \frac{1}{j}\leq1+\log k; \qquad \sum_{j=1}^k \frac{1}{j^2} \leq 2.$

Thus, when Re(α) ≤ -1, we have

$\left|{\alpha \choose k} \right|^2\leq \mathrm{e}^{- 2(1+\mathrm{Re}\,\alpha )(1+\log k) +2|1+\alpha|^2}$

When -1 ≤ Re(α), we have

$\left|{\alpha \choose k} \right|^2\leq \mathrm{e}^{- 2(1+\mathrm{Re}\,\alpha )\log k +2|1+\alpha|^2}$

Therefore, no matter Re(α) ≤ -1 or not, we always have

$\left|{\alpha \choose k} \right|^2\leq \frac{M^2}{k^{2(1+\mathrm{Re}\,\alpha )} }$

with a M independent of k, and prove the claim.

See also

• Binomial theorem
• Table of Newtonian series