Binomial series

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, where Re(α) denotes the real part of α.
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 either Re(α) > 0 or α = 0.
5. If |x| > 1, the series diverges, unless α is a non-negative integer (in which case the series is a finite sum).

In particular, if ${\displaystyle \alpha }$ is not a non-negative integer, the situation at the boundary of the disk of convergence, ${\displaystyle |x|=1}$, is summarized as follows:

• 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.

The following hold for any complex number α:

${\displaystyle {\alpha \choose 0}=1,}$

${\displaystyle {\alpha \choose k+1}={\alpha \choose k}\,{\frac {\alpha -k}{k+1}},}$

(2)

${\displaystyle {\alpha \choose k-1}+{\alpha \choose k}={\alpha +1 \choose k}.}$

(3)

Unless ${\displaystyle \alpha }$ is a nonnegative integer (in which case the binomial coefficients vanish as ${\displaystyle k}$ is larger than ${\displaystyle \alpha }$), a useful asymptotic relationship for the binomial coefficients is, in Landau notation:

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

(4)

This is essentially equivalent to Euler's definition of the Gamma function:

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

and implies immediately the coarser bounds

${\displaystyle {\frac {m}{k^{1+\operatorname {Re} \,\alpha }}}\leq \left|{\alpha \choose k}\right|\leq {\frac {M}{k^{1+\operatorname {Re} \alpha }}},}$

(5)

for some positive constants m and M .

Formula (2) for the generalized binomial coefficient can be rewritten as

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

(6)

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

${\displaystyle \sum _{k=1}^{\infty }\;{\frac {1}{k^{p}}},}$

with ${\displaystyle p=1+\operatorname {Re} \alpha }$. To prove (iii), first use formula (3) to obtain

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

(7)

and then use (ii) and formula (5) again to prove convergence of the right-hand side when ${\displaystyle \operatorname {Re} \alpha >-1}$ is assumed. On the other hand, the series does not converge if ${\displaystyle |x|=1}$ and ${\displaystyle \operatorname {Re} \alpha \leq -1}$, again by formula (5). Alternatively, we may observe that for all ${\displaystyle j}$, ${\textstyle \left|{\frac {\alpha +1}{j}}-1\right|\geq 1-{\frac {\operatorname {Re} \alpha +1}{j}}\geq 1}$. Thus, by formula (6), for all ${\textstyle k,\left|{\alpha \choose k}\right|\geq 1}$. This completes the proof of (iii). Turning to (iv), we use identity (7) above with ${\displaystyle x=-1}$ and ${\displaystyle \alpha -1}$ in place of ${\displaystyle \alpha }$, along with formula (4), to obtain

${\displaystyle \sum _{k=0}^{n}\;{\alpha \choose k}\;(-1)^{k}={\alpha -1 \choose n}\;(-1)^{n}={\frac {1}{\Gamma (-\alpha +1)n^{\alpha }}}(1+o(1))}$

as ${\displaystyle n\to \infty }$. Assertion (iv) now follows from the asymptotic behavior of the sequence ${\displaystyle n^{-\alpha }=e^{-\alpha \log(n)}}$. (Precisely, ${\displaystyle \left|e^{-\alpha \log n}\right|=e^{-\operatorname {Re} \alpha \,\log n}}$ certainly converges to ${\displaystyle 0}$ if ${\displaystyle \operatorname {Re} \alpha >0}$ and diverges to ${\displaystyle +\infty }$ if ${\displaystyle \operatorname {Re} \alpha <0}$. If ${\displaystyle \operatorname {Re} \alpha =0}$, then ${\displaystyle n^{-\alpha }=e^{-i\operatorname {Im} \alpha \log n}}$ converges if and only if the sequence ${\displaystyle \operatorname {Im} \alpha \log n}$ converges ${\displaystyle {\bmod {2\pi }}}$, which is certainly true if ${\displaystyle \alpha =0}$ but false if ${\displaystyle \operatorname {Im} \alpha \neq 0}$: in the latter case the sequence is dense ${\displaystyle {\bmod {2\pi }}}$, due to the fact that ${\displaystyle \log n}$ diverges and ${\displaystyle \log(n+1)-\log n}$ converges to zero).