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edited Dec 4, 2018 at 3:06

Mark Girard

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

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asked Dec 4, 2018 at 2:12

Mark Girard

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

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.

sequences-and-series limits-and-convergence