Timeline for Strange limit problem involving $\binom{z}{n} e^{-xn\log n}$ with $z \in \mathbb{C}$

8 events

when toggle format	what		by	license	comment
Feb 3, 2016 at 18:42	vote	accept	r9m
Feb 3, 2016 at 18:31	answer	added	Fedor Petrov		timeline score: 5
Feb 3, 2016 at 17:37	comment	added	Fedor Petrov		If $\Re z>0$, the series $\sum_n \binom{z}n$ converges absolutely (by Raabe test, for example), thus we may replace to 0 and just need to check that the sum equals $2^z$. This follows from another relaxation: $\sum \binom{z}{n}x^n=(1+x)^z$ for $0<x<1$, and both parts are continuous at point 1.
Feb 3, 2016 at 16:38	comment	added	r9m		@CarloBeenakker I understand the intuition, infact for $z = -1$ I split the sum into odd and even terms, replaced them with integrals and a limit theorem ensured that we can replace $e^{-xn\log n}$ with $e^{-nx}$ without changing the limit, but not sure how to proceed for other cases of $z$. Say for $z = -\frac{1}{2}$ I am having troubles establishing the same result.
Feb 3, 2016 at 16:30	comment	added	Carlo Beenakker		this factor $n^{-xn}$ with $x$ a positive infinitesimal allows you to take the limit $y\rightarrow 1$
Feb 3, 2016 at 16:29	comment	added	Carlo Beenakker		isn't this just Newton's generalized binomial theorem $(1+y)^z=\sum_{n=0}^\infty {z\choose n} y^n$, $\|y\|<1$, with a convergence factor to allow for $y=1$?
Feb 3, 2016 at 16:28	comment	added	r9m		@CarloBeenakker I understand that but what do you propose I do with $\log n$ part in $e^{-x n\log n}$?
Feb 3, 2016 at 16:00	history	asked	r9m	CC BY-SA 3.0