# Binomial series

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I 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{aligned}(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{aligned}} 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 converges if and only if it converges absolutely.
5. 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{aligned}{\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{aligned}} 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

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

Using formula (2), it is easy to prove by induction that

${\alpha \choose k}=\prod _{j=1}^{k}\left({\frac {\alpha +1}{j}}-1\right).\qquad \qquad (6)$ ### 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

$\sum _{k=1}^{\infty }{\frac {1}{k^{p}}},$ with p = 1 + Re(α).

To prove (iii), first use formula (3) to obtain

$(1+x)\sum _{k=0}^{n}{\binom {\alpha }{k}}x^{k}=\sum _{k=0}^{n}{\binom {\alpha +1}{k}}x^{k}+{\binom {\alpha }{n}}x^{n+1},$ 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 j,

$\left|{\frac {\alpha +1}{j}}-1\right|\geq 1-{\frac {\operatorname {Re} \alpha +1}{j}}\geq 1.$ Thus, by formula (6),

$\left|{\binom {\alpha }{k}}x^{k}\right|\geq 1.$ (Formula (5) also implies divergence.)

For (iv), first use the identity above, for x = −1 and with α - 1 in place of α to obtain

$\sum _{k=0}^{n}{\binom {\alpha }{k}}(-1)^{k}={\binom {\alpha -1}{n}}(-1)^{n}.$ If Re(α) < 0, formula (5) implies that the series diverges. Now assume that Re(α) = 0 but α ≠ 0. We have from formula (6) that

${\binom {\alpha -1}{n}}(-1)^{n}=\prod _{k=1}^{n}\left(1-{\frac {\alpha }{k}}\right).$ Suppose that this product sequence converges. Using formula (5) again, we see that it cannot converge to 0. It follows that the series

$\sum _{k=1}^{\infty }\log \left(1-{\frac {\alpha }{k}}\right)$ converges. Now using the Mercator series for the principal branch of the complex logarithm, we obtain

$\log \left(1-{\frac {\alpha }{k}}\right)+{\frac {\alpha }{k}}=O\left({\frac {1}{k^{2}}}\right){\text{ as }}k\to \infty .$ (See Big O notation.) This implies that the series

$\sum _{k=1}^{\infty }\left(\log \left(1-{\frac {\alpha }{k}}\right)+{\frac {\alpha }{k}}\right)$ converges (absolutely). It follows that $\sum _{k=1}^{\infty }{\frac {1}{k}}$ converges, which is false, showing that the product sequence above must in fact diverge. This completes the proof of (iv).

## 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) = αu(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 ck 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.[citation needed] 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, we shall derive the asymptotics (5) by proving the inequality

${\frac {m}{k^{\operatorname {Re} \alpha }}}\leq \left|{\alpha -1 \choose k}\right|\leq {\frac {M}{k^{\operatorname {Re} \alpha }}},{\text{ for all }}k\geq 1,$ with $0 and with m and M independent of k. Of course, we must assume that α is not a positive integer.

First, using formula (6) and a straightforward computation with complex numbers, note that

$\left|{\alpha -1 \choose k}\right|^{2}=\prod _{j=1}^{k}\left|{\frac {\alpha }{j}}-1\right|^{2}=\prod _{j=1}^{k}\left(1-{\frac {2\operatorname {Re} \alpha }{j}}+\left({\frac {|\alpha |}{j}}\right)^{2}\right).$ Now using the Mercator series for the logarithm, we obtain

$\log \left(1-{\frac {2\operatorname {Re} \alpha }{j}}+\left({\frac {|\alpha |}{j}}\right)^{2}\right)+{\frac {2\operatorname {Re} \alpha }{j}}=O({\frac {1}{j^{2}}}){\text{ as }}j\to \infty$ .

This implies that the series

$\sum _{j=1}^{\infty }\left(\log \left(1-{\frac {2\operatorname {Re} \alpha }{j}}+\left({\frac {|\alpha |}{j}}\right)^{2}\right)+{\frac {2\operatorname {Re} \alpha }{j}}\right)$ converges (absolutely), so its partial sums are certainly bounded, whence

$\log \left|{\alpha -1 \choose k}\right|+\sum _{j=1}^{k}{\frac {\operatorname {Re} \alpha }{j}}$ must be a bounded sequence. (Since α is not a positive integer, we are never taking the log of 0.)

From the definition of the natural logarithm it is easily shown that

$\log k<\log(k+1)<\sum _{j=1}^{k}{\frac {1}{j}}<1+\log k,$ from which it follows that the sequence

$(\operatorname {Re} \alpha )\log k-\sum _{j=1}^{k}{\frac {\operatorname {Re} \alpha }{j}}$ is bounded. Thus, the sequence

$\log \left|{\alpha -1 \choose k}\right|+(\operatorname {Re} \alpha )\log k$ must be bounded. From this it is clear that we can find m and M independent of k such that

$0 This completes the proof.