Timeline for Alternating sum of square roots of binomial coefficients

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12 events

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Apr 13, 2017 at 12:58	history	edited	CommunityBot		replaced http://mathoverflow.net/ with https://mathoverflow.net/
Mar 11, 2012 at 18:59	history	edited	Mark Wildon		Added binomial-coefficient tag
Jan 6, 2012 at 18:21	comment	added	GH from MO		Noam's argument even gives $\sum_{r=0}^n {n\choose r}^\alpha a^r b^{n-r}\geq 0$ for any real $a$ and $b$, even $n$, and $\alpha\leq 1$. See my comment to his response.
Jan 6, 2012 at 18:12	comment	added	Roland Bacher		$n\longmapsto \sum_{r=0}^{2n}(-1)^r{2n\choose r}^\alpha$ seems to be bounded for all real $\alpha\leq 3/2$ and seems to be unbounded for $\alpha>3/2$.
Jan 6, 2012 at 14:55	vote	accept	Mark Wildon
Jan 6, 2012 at 10:21	comment	added	Roland Bacher		A related (and fairly easy) fact is $\lim_{n\rightarrow\infty}\sum_{k=0}^{2n}(-1)^k{2n\choose k}^{1/{2n\choose k}}=1$.
Jan 6, 2012 at 6:53	answer	added	Noam D. Elkies		timeline score: 46
Jan 6, 2012 at 6:18	comment	added	GH from MO		Indeed $c_n>0$, read my response below. In a similar fashion $c_n>c_{n+2}$ should follow, too.
Jan 6, 2012 at 5:55	answer	added	GH from MO		timeline score: 32
Jan 6, 2012 at 2:06	comment	added	J Russell		Mark, yes, the positivity of this infinite set of finite sums feels to me quite similar to (and, as you point out, implies) the positivity of the infinite series I asked about earlier in the question you reference, and suffers from the same delicate cancellation of huge quantities. I sort of suspect that if you could crack the infinite series, you could crack this, too.
Jan 6, 2012 at 0:51	comment	added	user6976		If you denote $\sum_{r=0}^n (-1)^r \binom{n}{r}^\alpha$ by $f(n,\alpha)$, then $f(2n,1)=0, f(2n,0)=1$, and $f(2n,\alpha)$ seems to be decreasing with $\alpha$ for every $n$. That fact (which is stronger than both conjectures you mentioned) may be more feasible.
Jan 6, 2012 at 0:20	history	asked	Mark Wildon	CC BY-SA 3.0