Rate of convergence of binomial series This is the binomial series:
$$(1+x)^\alpha = \sum_{k=0}^\infty \binom{\alpha}{k} x^k$$
where $|x|<1$ and $\alpha$ can be a complex number in general. How fast does it converge? 
I need an upper bound of the error $E_n(x)$:
$$E_n(x) = \sum_{k=n+1}^\infty \binom{\alpha}{k} x^k$$
Specifically, I need a nice analytical estimate of $E_n(x)$ that I can use to predict how many terms of the series I will need in a given circumstance. I know that $E_n(x)$ can be expressed in terms of hypergeometric functions, but that's not very useful for numerical analysis of convergence. Is there a nicer estimate?
 A: If you intend to compute the answer to arbitrary precision, you would be far better off using $(1+x)^a = e^{a\ln(1+x)}$, since there are fast algorithms for computing elementary functions (see http://en.wikipedia.org/wiki/Computational_complexity_of_mathematical_operations and http://www.math.ust.hk/~machiang/education/enhancement/arithmetic_geometric.pdf).
Nevertheless, you can get simple bounds on the binomial coefficient by the following:
Let $c=\lceil a \rceil$
If $k \le c$,
  $\left| \binom{a}{k} \right| = \prod_{m=1}^k \frac{a+1-m}{m} \le \prod_{m=1}^k \frac{c+1-m}{m} = \binom{c}{k} \le \binom{c}{c/2} < {\large \frac{2^c}{\sqrt{\pi c/2}} }$
  (see http://en.wikipedia.org/wiki/Central_binomial_coefficient)
If $k > c$,
  $\left| \binom{a}{k} \right| = \prod_{m=1}^c \frac{a+1-m}{m} \times \prod_{m=c+1}^k \frac{m-(a+1)}{m}$
  $ \le \prod_{m=1}^c \frac{c+1-m}{m} \times \prod_{m=c+1}^k \frac{m-c}{m}$
  $ = 1 \times \prod_{m=c+1}^k \sqrt{\frac{(m-c)(k+1-m)}{m(k+c+1-m)}}$
  $ \le \left( \frac{k+1-c}{k+1+c} \right)^{k-c}$ because $\frac{(p-r)(p+r)}{(q-r)(q+r)} \le \frac{p^2}{q^2}$ for any $0 \le p \le q$ and $0 \le r < q$
These bounds are by no means tight, but good enough for their simplicity. Note that when $k>c$ the terms in your series are strictly decreasing in absolute value and alternating in sign and so you can use the bound for a particular term to bound the entire tail. But that is only if you have no choice but to use the series, since using argument reduction and the Taylor expansions for the exponential and logarithm will yield an algorithm that converges much faster (see http://www.jstor.org/discover/10.2307/2008657?uid=3738992&uid=2&uid=4&sid=21103354244373 or http://www.ams.org/mcom/1989-52-185/S0025-5718-1989-0971406-0/S0025-5718-1989-0971406-0.pdf if you don't have access to JSTOR).
A: The binomial coefficient can be expressed in terms of Gamma functions, from which you can easily get its asymptotics -- Mathematica gets the following ($x$ is your $\alpha$) From this you see that the binomials go to zero like $1/k.$
$$
\sin (\pi  (-k+x+1)) \left(\frac{\Gamma (x+1) \sqrt{\frac{1}{k}}}{\sqrt{2}
   \pi ^{3/2}}-\frac{\Gamma (x+1) \left(\frac{1}{k}\right)^{3/2}}{12
   \left(\sqrt{2} \pi ^{3/2}\right)}+\frac{\Gamma (x+1)
   \left(\frac{1}{k}\right)^{5/2}}{288 \sqrt{2} \pi
   ^{3/2}}+O\left(\left(\frac{1}{k}\right)^{7/2}\right)\right) \exp
   \left(\frac{1}{2} \left(2 x \log \left(\frac{1}{k}\right)+\log
   \left(\frac{1}{k}\right)+\log (2 \pi )\right)+\frac{6 x^2+6 x+1}{12
   k}+\frac{2 x^3+3 x^2+x}{12
   k^2}+O\left(\left(\frac{1}{k}\right)^3\right)\right)$$
