Editing Binomial series (section)

== Convergence ==
=== Conditions for convergence ===

Whether ({{EquationNote|1}}) [[convergent series|converges]] depends on the values of the complex numbers {{mvar|α}} and&nbsp;{{mvar|x}}. More precisely:

#If {{math|{{!}}''x''{{!}} < 1}}, the series converges [[absolute convergence|absolutely]] for any complex number {{mvar|α}}. 
#If {{math|1={{!}}''x''{{!}} = 1}}, the series converges absolutely [[if and only if]] either {{math|Re(''α'') > 0}} or {{math|1=''α'' = 0}}, where {{math|Re(''α'')}} denotes the [[complex number|real part]] of {{mvar|α}}.
# If {{math|1={{!}}''x''{{!}} = 1}} and {{math|''x'' ≠ −1}}, the series converges if and only if {{math|Re(''α'') > −1}}.
#If {{math|1=''x'' = −1}}, the series converges if and only if either {{math|Re(''α'') > 0}} or {{math|1=''α'' = 0}}.
#If {{math|{{!}}''x''{{!}} > 1}}, the series [[divergent series|diverges]] except when {{mvar|α}} is a non-negative integer, in which case the series is a finite sum.

In particular, if {{mvar|α}} is not a non-negative integer, the situation at the boundary of the [[radius of convergence|disk of convergence]], {{math|1={{abs|''x''}} = 1}}, is summarized as follows:

* If {{math|Re(''α'') > 0}}, the series converges absolutely.
* If {{math|−1 < Re(''α'') ≤ 0}}, the series converges [[conditional convergence|conditionally]] if {{math|''x'' ≠ −1}} and diverges if {{math|1=''x'' = −1}}.
* If {{math|Re(''α'') ≤ −1}}, the series diverges.

=== Identities to be used in the proof ===

The following hold for any complex number&nbsp;{{mvar|α}}:

:<math>{\alpha \choose 0} \!= 1,</math>
{{NumBlk|:|<math> {\alpha \choose k+1} \!= \!{\alpha\choose k}  \frac{\alpha-k}{k+1}, </math>|{{EquationRef|2}}}}

{{NumBlk|:|<math> {\alpha \choose k-1} \!+ \!{\alpha\choose k} \!= \!{\alpha+1 \choose k}. </math>|{{EquationRef|3}}}}
Unless <math>\alpha</math> is a nonnegative integer (in which case the binomial coefficients vanish as <math>k</math> is larger than <math>\alpha</math>), a useful [[asymptotic analysis|asymptotic]] relationship for the binomial coefficients is, in [[Landau notation]]:

{{NumBlk|:|<math> {\alpha \choose k} \!= \frac{(-1)^k} {\Gamma(-\alpha)k^ {1+\alpha} } \,(1+o(1)), \quad\text{as }k\to\infty. </math>|{{EquationRef|4}}}}

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

:<math>\Gamma(z) = \lim_{k \to \infty} \frac{k! \,k^z}{z(z+1)\cdots(z+k)}, </math>

and implies immediately the coarser bounds

{{NumBlk|:|<math> \frac {m} {k^{1+\operatorname{Re}\,\alpha}}\le \left|{\alpha \choose k}\right| \le \frac {M} {k^{1+\operatorname{Re}\alpha}}, </math>|{{EquationRef|5}}}}
for some positive constants {{mvar|m}} and {{mvar|M}} .

Formula ({{EquationNote|2}}) for the generalized binomial coefficient can be rewritten as
{{NumBlk|:|<math> {\alpha \choose  k} \!= \prod_{j=1}^k \!\left(\frac{\alpha + 1}j - 1 \right). </math>|{{EquationRef|6}}}}

=== Proof ===

To prove (i) and (v), apply the [[ratio test]] and use formula ({{EquationNote|2}}) above to show that whenever <math>\alpha</math> is not a nonnegative integer, the [[radius of convergence]] is exactly&nbsp;1.  Part (ii) follows from formula ({{EquationNote|5}}), by comparison with the [[Convergence tests#p-series test|{{mvar|p}}-series]]

:<math> \sum_{k=1}^\infty \frac1{k^p}, </math>

with <math>p=1+\operatorname{Re}(\alpha)</math>. To prove (iii), first use formula ({{EquationNote|3}}) to obtain

{{NumBlk|:|<math>(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}, </math>|{{EquationRef|7}}}}

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

:<math>\sum_{k=0}^n \!{\alpha\choose k} (-1)^k = \!{\alpha-1 \choose n} (-1)^n= \frac1{\Gamma(-\alpha+1)n^\alpha} (1+o(1))</math>

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