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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)?

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  • $\begingroup$ Your computation of ${n\choose 2}$ is incorrect. $\endgroup$ – Dietrich Burde Mar 7 '15 at 19:36
  • $\begingroup$ Indeed. $\binom{n}{2} = \frac{n(n-1)}{2}$ and $\binom{n-1}{2} = \frac{(n-1)(n-2)}{2}$, leading you for the center line to be $\lim \frac{1}{2}\cdot \frac{n}{n-2}<1$ so it converges. $\endgroup$ – JMoravitz Mar 7 '15 at 19:37
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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
$$ \sum_{n=12}^\infty \frac{(n-1)(n-2)}{2}\;\left(\frac{1}{2}\right)^n = \frac{67}{2048} \approx 0.0327 $$

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  • $\begingroup$ Ok, I understand this solution, but what if $\sum (n-1)(n-2)..(n-5)(\frac{1}{2})^n$. Should I differentiate 5 times? $\endgroup$ – Lil'Lobster Mar 7 '15 at 19:54
  • $\begingroup$ That generalization is left to the reader. $\endgroup$ – GEdgar Mar 7 '15 at 19:55
  • $\begingroup$ So I kindly ask for a short, emboldening hint ("yes"/ "no"). $\endgroup$ – Lil'Lobster Mar 7 '15 at 20:00
  • $\begingroup$ @Lili: Yes. $\,$ $\endgroup$ – Brian M. Scott Mar 7 '15 at 20:12
  • $\begingroup$ @Lili, why not just try it and see what happens?! $\endgroup$ – Mariano Suárez-Álvarez Mar 7 '15 at 20:17

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