Timeline for Lower bound and limit of a sum with binomial coefficients

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Apr 13, 2022 at 5:53	comment	added	Jorge Zuniga		This means that $T_{n+1}>γ(n)⋅T_n$ holds for all $n>\max(\nu)+3=4.12596...,$, namely from $n=n_0=5$ and claim is proved.
Apr 12, 2022 at 17:52	comment	added	Jorge Zuniga		@macat checked. The right-most real roots are respectively at $\nu=[1.12596.., 1.11566..]$ for [numerator, denominator] of rational function $[1+\alpha]\,r_1+r_2+r_3$
Apr 11, 2022 at 6:20	history	edited	Jorge Zuniga	CC BY-SA 4.0	edited body
Apr 10, 2022 at 20:48	comment	added	Jorge Zuniga		@macat, OK. If the first $T_n>0$ is at $n_0=5$ you have to prove that this rational $[1+\alpha]\,r_1+r_2+r_3>0$ from this value. This is a ratio of two polynomials in $n$. Look for the real roots of the resulting numerator polynomial first and then the real roots of the denominator polynomial. None polynomial should have real roots to the right of $n_0$ and both must have same sign for all $n>n_0$. $\ T_{n+1}>\gamma(n)\cdot T_n$ follows by induction from $n_0$. I do not think there is a pen-and-paper proof. You have to use a computer to get the roots.
Apr 9, 2022 at 23:11	comment	added	macat		The last proof by induction seems to work only from $n_0=5$, because we need that the $T_n$'s are non-negative. Is there a chance that this approach can be turned into a pen-and-paper proof?
Apr 5, 2022 at 19:05	history	answered	Jorge Zuniga	CC BY-SA 4.0