how to calculate sum of a series? (me or Wangenmakers is wrong)

Question

Wagenmakers in his critical article about p-values wrote that:

$$\sum_{i=12}^{\infty} {{n-1} \choose {2}} \cdot \left(\frac{1}{2}\right)^n \approx .033$$

How could he do his calculations if the D'Alembert's criterion shows that the series diverges?

$\lim_{n\rightarrow \infty} \frac{a_{n+1}}{a_n}=\lim_{n\rightarrow \infty}\frac{ {n \choose 2} \cdot (\frac{1}{2})^{n+1} } {{{n-1} \choose {2}}\cdot(\frac{1}{2})^n}= \lim_{n\rightarrow \infty} \frac{\frac{n!(n-2)!}{2!}}{\frac{(n-1)!\cdot(n-3)!}{2!}}\cdot\frac{1}{2}=\lim_{n\rightarrow\infty}n\cdot (n-2)\cdot\frac{1}{2}=\infty >1$

If I'm wrong, how can I evaluate the series above explicitly (to reach this 0.033)?

Answer 1

Exact... First, $$ \sum_{n=12}^\infty \frac{(n-1)(n-2)}{2}\;x^n = \frac{x^{12}(45x^2-99x+55)}{(1-x)^3} $$ obtained by differentiating the geometric series $\sum_{n=12}^\infty x^{n-1}$ twice.
So
$$ \sum_{n=12}^\infty \frac{(n-1)(n-2)}{2}\;\left(\frac{1}{2}\right)^n = \frac{67}{2048} \approx 0.0327 $$