Timeline for Value of $c$ such that $\lim_{n\rightarrow\infty}\sum_{k=1}^{n-1}\frac{1}{(n-k)c+\log(n!)-\log(k!)}=1$

8 events

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Dec 13, 2018 at 19:00	comment	added	esg		For the lower bound write $S(n,c)=\sum_{k=1}^{n-1}\frac{1}{k\,c+\log(n!)-\log((n-k)!)}$ and use $\log(n!)-\log((n-k)!)\leq k\log(n)$ to find that $$S(n,c)\geq \frac{H_{n-1}}{c+\log(n)}\;\;,$$ where $H_{n-1}$ is the $(n-1)-$th harmonic number. Thus $\liminf S(n,c)\geq 1$ (independent of $c\geq 0$), as originally asserted.
Dec 5, 2018 at 10:24	comment	added	Yaakov Baruch		For what it is worth, I would recommend against deleting.
Dec 5, 2018 at 9:57	history	edited	Carlo Beenakker	CC BY-SA 4.0	added 482 characters in body
Dec 5, 2018 at 9:38	history	edited	Carlo Beenakker	CC BY-SA 4.0	deleted 234 characters in body
Dec 4, 2018 at 22:25	history	edited	Carlo Beenakker	CC BY-SA 4.0	added 124 characters in body
Dec 4, 2018 at 22:19	history	edited	Carlo Beenakker	CC BY-SA 4.0	added 345 characters in body
Dec 4, 2018 at 13:10	history	edited	Carlo Beenakker	CC BY-SA 4.0	added 287 characters in body
Dec 4, 2018 at 10:06	history	answered	Carlo Beenakker	CC BY-SA 4.0