Show that $1+\sum_{k=1}^\infty\left\{1-\left(k+\frac12\right)\log\left(\frac{k+1}{k}\right)\right\}=\frac12\times\log(2\pi)$ I tried to prove that the constant in Stirling's formula is $\sqrt {2\pi}$. Then I used the Euler-Maclaurin formula, so finally this is what I need to prove. Wikipedia gives a hint that we need to use Wallis' product, but I couldn't prove it. Please help.
 A: Let's define two sequences, $ \left(a_{n}\right)_{n\geq 0} , \left(b_{n}\right)_{n>0}\in\mathbb{R}^{\mathbb{N}} $ as follows : $$ \left(\forall n\in\mathbb{N}\right),\ a_{n}=\frac{n^{n}\mathrm{e}^{-n}\sqrt{n}}{n!}\\ \left(\forall n\in\mathbb{N}^{*}\right),\ b_{n}=\ln{\left(\frac{a_{n+1}}{a_{n}}\right)} $$
Let $ n\in\mathbb{N}^{*} $, we have $ \left(\forall k\in\mathbb{N}^{*}\right) $, the following : \begin{aligned} b_{k}&=\ln{\left(\left(\frac{k+1}{k}\right)^{k+\frac{1}{2}}\mathrm{e}^{-1}\right)}\\ &=-1+\left(k+\frac{1}{2}\right)\ln{\left(1+\frac{1}{k}\right)} \end{aligned}
Thus : \begin{aligned} 1+\sum_{k=1}^{n-1}{\left(1-\left(k+\frac{1}{2}\right)\ln{\left(1+\frac{1}{k}\right)}\right)}&=1+\sum_{k=1}^{n-1}{\left(\ln{\left(a_{k+1}\right)}-\ln{\left(a_{k}\right)}\right)}\\ &=1+\ln{\left(a_{n}\right)}-\ln{\left(a_{1}\right)}\\ &=\ln{\left(a_{n}\right)} \end{aligned}
Since : $$ 1-\left(n+\frac{1}{2}\right)\ln{\left(1+\frac{1}{n}\right)}=\underset{\overset{n\to +\infty}{}}{\mathcal{O}}\left(\frac{1}{n^{2}}\right) $$
Then the series $ \sum\limits_{n\geq 1}{\left(1-\left(n+\frac{1}{2}\right)\ln{\left(1+\frac{1}{n}\right)}\right)} $ converges, which means that $ \ln{\left(a_{n}\right)}\underset{n\to +\infty}{\longrightarrow}\ell_{0}\in\mathbb{R} $, hence, denoting $ \ell=\mathrm{e}^{\ell_{0}} $, we have : $$ n!\underset{n\to +\infty}{\sim}\ell\sqrt{n}\left(\frac{n}{\mathrm{e}}\right)^{n} \ \ \ \ \fbox{$\begin{array}{rcl}1\end{array}$}$$
Now we just need to find what $ \ell $ is, to do so we'll be using the fact that : $$ \frac{}{}\frac{2}{\left(2n+1\right)\pi}\left(\frac{2^{2n}\left(n!\right)^{2}}{\left(2n\right)!}\right)^{2}=\frac{W_{2n+1}}{W_{2n}}\underset{n\to +\infty}{\longrightarrow}1  \ \ \ \ \fbox{$\begin{array}{rcl}2\end{array}$}$$
Were $ W_{n} $, for $ n\in\mathbb{N} $, stands for Wallis integral : $$ \left(\forall n\in\mathbb{N}\right),\ W_{n}=\int_{0}^{\frac{\pi}{2}}{\sin^{n}{x}\,\mathrm{d}x} $$
I'm not gonna detail the second result, simply because it's easy to prove, also it is an exercise that, I think, has been solved many times in this website.
Using our first result, we can write that : $$ \frac{2}{\left(2n+1\right)\pi}\left(\frac{2^{2n}\left(n!\right)^{2}}{\left(2n\right)!}\right)^{2}\underset{n\to +\infty}{\sim}\frac{2^{4n}}{n\pi}\times\frac{\ell^{4}n^{4n+2}\mathrm{e}^{-4n}}{\ell^{2}\left(2n\right)^{4n+1}\mathrm{e}^{-4n}}\underset{n\to +\infty}{\sim}\frac{\ell^{2}}{2\pi} $$
Thus $ \ell = \sqrt{2\pi} $, which means $ \ell_{0} $, which is the sum of our series, would then be equal to $ \frac{1}{2}\ln{\left(2\pi\right)} $.
