Show that $\log n! - (n+1/2)\log n + n$ has a finite limit when $n\to\infty$ I am trying to show $\lim\limits_{n \to \infty} (\log n! - (n+1/2)\log n + n)$ exists as part of a much bigger problem. I could really use some help.
My attempt: I tried to use the fact that $1+ \frac {1}{2}+ \frac {1}{3}+ ... +\frac {1}{n} - \log n$ exists, but I got nowhere.
 A: Hint. If you set
$$
u_n:=\log n! - (n+1/2)\log n + n,\qquad n=1,2,3,\cdots,
$$ then you may prove that

$$
u_{n+1}-u_n=1-\left(\frac{1}{2}+n\right) \log\left(1+\frac{1}{n}\right)\leq0 \tag1
$$ 

implying the sequence $\left\{u_n\right\}$ is decreasing. On the other hand, you have

$$
u_n\geq0. \tag2
$$ 

Thus $\lim\limits_{n \to \infty}u_n=\lim\limits_{n \to \infty} (\log n! - (n+1/2)\log n + n)$ exists.

Edit. Let's give some details.


*

*Proof of $(1)$. Setting $\displaystyle f(x):=\frac{2x}{x+2}-\log(1+x),\, x  \in [0,1],$ we have $$ f(0)=0, \quad
   f'(x)=\frac{-x^2}{(x+2)^2(x+1)}\leq0 \quad  \implies f(x)\leq0, \quad
   x \in [0,1], $$ giving, for $n\geq1$,  $$
   u_{n+1}-u_n=\left(\frac12+n\right)\underbrace{\color{#C00000}{\left(\frac1{\frac12+n}
   -\log\left(1+\frac{1}{n}\right)\right)}}_{\large \color{#C00000}{f\left(\frac1n
   \right)} \leq\:\color{#C00000}{0}} \leq \color{#C00000}{0} .$$

*Proof of $(2)$. One may observe that, for $n\geq1$,$$
\begin{align}
&u_n=\log n! - (n+1/2)\log n + n\\
&=\sum_{k=1}^n\log k-\int_{\frac12}^{n+\frac12}\log x\,\mathrm{d}x+(n+1/2)\log \left(1+\frac1{2n}\right)+\frac12\log 2\\
&=\underbrace{-\sum_{k=1}^nk\int_0^{\frac1{2k}}\log\left(1-x^2\right)\,\mathrm{d}x}_{\large \color{blue}{\geq \:0}}+\underbrace{(n+1/2)\log \left(1+\frac1{2n}\right)+\frac12\log 2}_{\large \color{blue}{\geq \:0}}\color{blue}{\geq 0}.
\end{align}
$$
A: Consider the following equation
$$
\begin{align}
&\color{#C00000}{\left(n+\tfrac12\right)\log\left(n+\tfrac12\right)-\left(n+\tfrac12\right)-\tfrac12\log\left(\tfrac12\right)+\tfrac12}\color{#00A000}{-\log(n!)}\\
&=\color{#C00000}{\int_{\frac12}^{n+\frac12}\log(x)\,\mathrm{d}x}\color{#00A000}{-\sum_{k=1}^n\log(k)}\\
&=\sum_{k=1}^n\int_{k-\frac12}^{k+\frac12}\log\left(\frac xk\right)\,\mathrm{d}x\\
&=\sum_{k=1}^nk\int_0^{\frac1{2k}}\log\left(1-x^2\right)\,\mathrm{d}x\\
&=\sum_{k=1}^nk\int_0^{\frac1{2k}}O\!\left(x^2\right)\,\mathrm{d}x\\
&=\sum_{k=1}^nk\,O\!\left(\frac1{k^3}\right)\\
&=\sum_{k=1}^nO\!\left(\frac1{k^2}\right)\\
&=C+O\!\left(\frac1n\right)\tag{1}
\end{align}
$$
Since the right side of $(1)$ converges as $n\to\infty$, so does the expression on the left.
From the expression on the left side of $(1)$,
$$
\begin{align}
\left(n+\tfrac12\right)\log\left(n+\tfrac12\right)
&=\left(n+\tfrac12\right)\left[\log(n)+\tfrac1{2n}+O\!\left(\tfrac1{n^2}\right)\right]\\[6pt]
&=\left(n+\tfrac12\right)\log(n)+\tfrac12+O\!\left(\tfrac1n\right)\tag{2}
\end{align}
$$
Combining $(1)$ and $(2)$, we get that
$$
\lim_{n\to\infty}\left[\left(n+\tfrac12\right)\log(n)-n-\log(n!)\right]=C-\tfrac12\log(2e)\tag{3}
$$
exists and is finite.
