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

Question

What is the value of $c$ such that $$\lim_{n\rightarrow\infty}\sum_{k=1}^{n-1}\frac{1}{(n-k)c+\log(n!)-\log(k!)}=1?$$ Numerically, it seems that the answer is $c=\log 2$. But I'd like to see a reason why.

Answer 1

UPDATE

I tried to evaluate the sum numerically for large $n$ and what I find does not support the conclusion I give below, that the large-$n$ limit equals 1 independent of $c$. Here is a plot for $c=\log 2$, obtained with Mathematica. I am uncertain about the numerical stability, but if this plot is to be trusted the $n\rightarrow\infty$ limit for $c=\log 2$ is below unity. Apologies for my mistaken conclusion, I might delete this answer, but perhaps it can still serve a purpose.

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One way to proceed is as follows. For large $n$ I approximate the sum $$S(c,n)=\sum_{k=1}^{n-1}\frac{1}{(n-k)c+\log(n!)-\log(k!)}$$ by an integral over $k$ and I approximate the factorials $n!$ and $k!$ by Stirling's formula, so instead of the sum $S(c,n)$ I evaluate the integral

$$I(c,n)=\int_{1}^{n-1} \frac{1}{(n+1/2)\ln n-(k+1/2)\ln k+(c-1) (n-k)}\,dk.$$

Now for the large-$n$ asymptotics of $I(c,n)$, I first note the upper bound $$I(c,n)\leq\int_{1}^{n-1} \frac{1}{(n+1/2)\ln n-(k+1/2)\ln n+(c-1) (n-k)}\,dk$$ $$\qquad\qquad=\frac{\ln (n-1)}{c-1+\ln n}\rightarrow 1\;\;\text{for}\;\;n\rightarrow\infty.$$

For a lower bound I expand the integrand around $k=n$, $$I(c,n)\geq \int_{1}^{n-1} \frac{1}{(n-k) \left(c+\frac{1}{2n}+\ln n\right)}\,dk$$ $$\qquad\qquad=\frac{2 n \ln (n-1)}{2 c n+2 n \ln n+1}\rightarrow 1\;\;\text{for}\;\;n\rightarrow\infty.$$

The plot shows $I(c,n)$ (blue) and the upper and lower bounds, for $c=\ln 2$ (at the left) and for $c=0.1$ (at the right). Notice that the lower bound is actually quite accurate already for moderately large $n$.

Answer 2

I think, that $c=\gamma=0.57721..$ (Euler-Mascheroni), because $$ \sum_{k=1}^{n-1} \frac{1}{(n-k) c + \ln \Gamma(n+1)-\ln \Gamma(k+1)} = \\ \sum_{m=1}^{n-1} \frac{1}{\ln(\frac{ e^{m c}\ \Gamma(n+1)}{\Gamma(n-m+1)})} \sim \sum_{m=1}^{n-1} \frac{1}{m \ \ln (e^{c}\ n)} = \\ \frac{1}{\ln (e^{c}\ n)} H_{n-1} \sim \frac{1}{\ln (e^{c}\ n)} (\ln n +\gamma) $$ where in the second line I changed the summation variable to $m=n-k$ and used $$ \frac{\Gamma(x+a)}{\Gamma(x+b)}\sim x^{a-b} $$ for large $x$.$H_n$ is the $n$-th Harmonic number and its asymptotic expansion for large $n$ is $$ H_{n}\sim \ln n +\gamma +O(n^{-1}). $$ All expansions/definitions can be found at Wikipedia or at https://dlmf.nist.gov/