Binomial series

where ${\displaystyle \alpha }$ is any complex number, and the power series on the right-hand side is expressed in terms of the (generalized) binomial coefficients

${\displaystyle (\genfrac{}{}{0ex}{}{\alpha }{k})=\frac{\alpha (\alpha -1)(\alpha -2)\cdots (\alpha -k+1)}{k!}.}$

The binomial series is the MacLaurin series for the function ${\displaystyle f(x)=(1+x{)}^{\alpha }}$. It converges when ${\displaystyle |x|<1}$.

If α is a nonnegative integer n then the x^{n + 1} term and all later terms in the series are 0, since each contains a factor of (n − n). In this case, the series is a finite polynomial, equivalent to the binomial formula.

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, 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 except when α is a non-negative integer, in which case the series is a finite sum.

In particular, if α is not a non-negative integer, the situation at the boundary of the disk of convergence, |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.

Identities to be used in the proof

The following hold for any complex number α:

${\displaystyle (\genfrac{}{}{0ex}{}{\alpha }{0})=1,}$

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:

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

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

and implies immediately the coarser bounds

for some positive constants m and M .

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

Proof

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

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

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

and then use (ii) and formula (5) again to prove convergence of the right-hand side when ${\displaystyle \mathrm{Re}(\alpha )>-1}$ is assumed. On the other hand, the series does not converge if ${\displaystyle |x|=1}$ and ${\displaystyle \mathrm{Re}(\alpha )\le -1}$, again by formula (5). Alternatively, we may observe that for all ${\displaystyle j}$, $\left|\frac{\alpha +1}{j}-1\right|\ge 1-\frac{\mathrm{Re}(\alpha )+1}{j}\ge 1$. Thus, by formula (6), for all $k,\left|(\genfrac{}{}{0ex}{}{\alpha }{k})\right|\ge 1$. 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

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

as ${\displaystyle n\to \mathrm{\infty }}$. Assertion (iv) now follows from the asymptotic behavior of the sequence ${\displaystyle n^{-\alpha }=e^{-\alpha \log (n)}}$. (Precisely, ${\displaystyle \left|e^{-\alpha \log n}\right|=e^{-\mathrm{Re}(\alpha )\log n}}$ certainly converges to ${\displaystyle 0}$ if ${\displaystyle \mathrm{Re}(\alpha )>0}$ and diverges to ${\displaystyle +\mathrm{\infty }}$ if ${\displaystyle \mathrm{Re}(\alpha )<0}$. If ${\displaystyle \mathrm{Re}(\alpha )=0}$, then ${\displaystyle n^{-\alpha }=e^{-i\mathrm{Im}(\alpha )\log n}}$ converges if and only if the sequence ${\displaystyle \mathrm{Im}(\alpha )\log n}$ converges ${\displaystyle mod2\pi }$, which is certainly true if ${\displaystyle \alpha =0}$ but false if ${\displaystyle \mathrm{Im}(\alpha )\ne 0}$: in the latter case the sequence is dense ${\displaystyle mod2\pi }$, due to the fact that ${\displaystyle \log n}$ diverges and ${\displaystyle \log (n+1)-\log n}$ converges to zero).