Interesting limit with log function Compute the following limit:$$ \lim _{n \to \infty} \frac{1}{n}\sum_{k=0}^{n-1}\left[\frac{k+1}{n}-\frac{1}{2}\right]\log(2n-2k-1) .$$ I do not know how to start as I am new in this subject. Can you help me?
 A: If we get an equivalent limit of the form $\lim_{n\to +\infty}\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{k}{n}\right)$ or $\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right)$ with $f$ being a Riemann-integrable function over $[0,1]$ we are done, since such a limit equals $\int_{0}^{1}f(x)\,dx$ that is hopefully simple to compute through the fundamental theorem of Calculus.
We may notice that $\sum_{k=0}^{n-1}\left(\frac{k+1}{n}-\frac{1}{2}\right)=\frac{1}{2}$, hence the given limit is equivalent to
$$ \lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}-\frac{1}{2}\right)\log\left(2-2\frac{k}{n}+\frac{1}{n}\right).$$
If we replace $\log\left(2-2\frac{k}{n}+\frac{1}{n}\right)$ with $\log\left(2-2\frac{k}{n}\right)$ the limit is still the same: I leave to you to prove this part through elementary inequalities. By the intro,
$$ \lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}-\frac{1}{2}\right)\log\left(2-2\frac{k}{n}\right)=\int_{0}^{1}\left(x-\frac{1}{2}\right)\log(2-2x)\,dx$$
equals:
$$ \int_{0}^{1}\left(x-\frac{1}{2}\right)\log(1-x)\,dx = \int_{0}^{1}\left(\frac{1}{2}-x\right)\log(x)\,dx =\color{red}{-\frac{1}{4}}$$
by differentiation under the integral sign or integration by parts.

As a consequence,
$$\lim_{n \to \infty} \frac {\sqrt[2n] {(2n-1)!!}} {\sqrt [n^2] a_n}=\color{red}{\large e^{1/4}}$$
as already explained here.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\lim _{n \to \infty}{1 \over n}\sum_{k = 0}^{n - 1}
\pars{{k + 1 \over n} - {1 \over 2}}\ln\pars{2n - 2k - 1}
\\[5mm] = &\
{1 \over 2}\lim _{n \to \infty}{1 \over n^{2}}\sum_{k = 0}^{n - 1}
\bracks{2\pars{n - 1 - k} + 2 - n}\ln\pars{2n - 2\bracks{n - 1 - k} - 1} =
\\[5mm] = &\
{1 \over 2}\lim _{n \to \infty}{1 \over n^{2}}\sum_{k = 0}^{n - 1}
\pars{n - 2k}\ln\pars{2k + 1} =
{1 \over 2}\lim _{n \to \infty}{n\sum_{k = 0}^{n - 1}\ln\pars{2k + 1} -
2\sum_{k = 0}^{n - 1}k\ln\pars{2k + 1} \over n^{2}}
\\[5mm] = &\
{1 \over 2}\lim _{n \to \infty}{\pars{n + 1}\sum_{k = 0}^{n}\ln\pars{2k + 1} -
n\sum_{k = 0}^{n - 1}\ln\pars{2k + 1} -
2n\ln\pars{2n + 1} \over \pars{n + 1}^{2} - n^{2}}\label{1}\tag{1}
\\[5mm] = &\
{1 \over 2}\lim _{n \to \infty}
{\sum_{k = 0}^{n}\ln\pars{2k + 1} - n\ln\pars{2n + 1} \over 2n + 1}
\label{2}\tag{2}
\\[5mm] = &\
{1 \over 2}\lim _{n \to \infty}
{\ln\pars{2n + 3} - \pars{n + 1}\ln\pars{2n + 3} + n\ln\pars{2n + 1} \over
\pars{2n + 3} - \pars{2n + 1}} =
-\,{1 \over 4}\lim_{n \to \infty}\bracks{n\ln\pars{2n + 3 \over 2n + 1}}
\label{3}\tag{3}
\\[5mm] = &\
-\,{1 \over 4}\ \underbrace{\lim_{n \to \infty}
\bracks{n\ln\pars{1 + {1 \over n + 1/2}}}}_{\ds{=\ 1}} =
\bbox[#ffe,15px,border:1px dotted navy]{\ds{-\,{1 \over 4}}}
\end{align}

In \eqref{1}, \eqref{2} and \eqref{3} I used the
  Stolz-Ces$\mrm{\grave{a}}$ro Theorem.

