Find $\lim_{n\to\infty}\frac{1}{n}\left(n+\frac{n-1}{2}+\frac{n-2}{3}+\dots+\frac{2}{n-1}+\frac{1}{n}-\log(n!)\right)$ I want to determine the limit of the following sequence
$$x_n=\frac{1}{n}\left(n+\frac{n-1}{2}+\frac{n-2}{3}+\dots+\frac{2}{n-1}+\frac{1}{n}-\log(n!)\right)$$
From the foregoing, consider
$$n+\frac{n-1}{2}+\frac{n-2}{3}+\dots+\frac{2}{n-1}+\frac{1}{n}=\sum_{k=1}^n\frac{n+1-k}{k}$$
Also try to consider Stirling's approximation, so you would have to find the limit of
$$x_n=\frac{1}{n}\left(\sum_{k=1}^n\frac{n+1-k}{k}-\log(\sqrt{2πn}\left(\frac{n}{e}\right)^n\right)$$
I don't know if my previous statement is completely true, besides, from this expression it is difficult for me to find the requested limit.
Any help please?
 A: Let $x_n$ be the sequence given by
$$\begin{align}
x_n&=\frac1n \left(\sum_{k=1}^n \frac{n+1-k}{k}-\log(k)\right)\\\\
&=\frac1n \left(\sum_{k=1}^n \frac nk+\frac1k-1-\log(k)\right)\\\\
&=-1+\underbrace{\frac1n\sum_{k=1}^n \frac1k}_{\to 0\,\,\text{as}\,\,n\to \infty} +\underbrace{\sum_{k=1}^n \frac1k-\log(n)}_{\to \gamma\,\,\text{as}\,\,n\to \infty}-\underbrace{\frac 1n\sum_{k=1}^n \log(k/n)}_{\to -1\,\,\text{as}\,\,n\to \infty}\tag1 
\end{align}$$
Therefore, we find that
$$\lim_{n\to \infty}x_n=\gamma$$
A: By the Stolz-Cesàro theorem (https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem),
\begin{eqnarray}
\lim_{n\to\infty}x_n&=&\lim_{n\to\infty}\frac1n \sum_{k=1}^n \left(\frac{n+1-k}{k}-\log(k)\right)\\
&=&\lim_{n\to\infty}\frac{\sum_{k=1}^n \left(\frac{n+1-k}{k}-\log(k)\right)}{n}\\
&=&\lim_{n\to\infty}\frac{\sum_{k=1}^{n+1} \left(\frac{n+2-k}{k}-\log(k)\right)-\sum_{k=1}^n \left(\frac{n+1-k}{k}-\log(k)\right)}{(n+1)-n}\\
&=&\lim_{n\to\infty}\sum_{k=1}^{n+1} \left(\frac{n+1-k}{k}+\frac1k-\log(k)\right)-\sum_{k=1}^n \left(\frac{n+1-k}{k}-\log(k)\right)\\
&=&\lim_{n\to\infty}\sum_{k=1}^{n+1}\frac1k+ \sum_{k=1}^{n+1}\left(\frac{n+1-k}{k}-\log(k)\right)-\sum_{k=1}^n \left(\frac{n+1-k}{k}-\log(k)\right)\\
&=&\lim_{n\to\infty}\sum_{k=1}^{n+1}\frac1k-\log(n+1)\\
&=&\gamma
\end{eqnarray}
