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

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Dec 6, 2018 at 15:08	comment	added	Carlo Beenakker		correct me if I'm wrong, but my impression is that this calculation expands the numerator to first order around $k=n$, to evaluate the sum $$\sum_{k=1}^{n-1}\left[\left(n-k) (c+\psi ^{(0)}(n+1)\right)\right]^{-1}=\frac{\gamma+\psi ^{(0)}(n)}{c+\psi ^{(0)}(n+1)},$$ with $\psi$ the polygamma function, which tends to 1 in the limit $n\rightarrow\infty$ irrespective of $c$.
Dec 5, 2018 at 23:10	comment	added	Yaakov Baruch		You seem to use $\frac{\Gamma(n-1)}{\Gamma(n-m+1)}\sim n^m$ for $1\le m \le n-1$. But $n-m$ (or even $n$) is not arbitrarily large relative to $m$ in most of that range. Notice that you reach the same (likely erroneous) conclusion that the limit is $1$ regardless of $c$, as in the previous answer.
Dec 5, 2018 at 14:01	history	edited	Johannes Trost	CC BY-SA 4.0	typo correction
Dec 5, 2018 at 13:00	history	answered	Johannes Trost	CC BY-SA 4.0