Prove $\sum_{n=1}^{\infty}((n+\frac{1}{2})\ln(1+\frac{1}{n})-1)=1-\ln(\sqrt{2\pi})$ I am looking for a derivation of the following sum:
$$\sum_{n=1}^{\infty}\bigg(\left(n+\frac{1}{2}\right)\ln\left(1+\frac{1}{n}\right)-1\bigg)=1-\ln(\sqrt{2\pi})$$
My current derivation(s) uses the zeta function at negative integers (and or Stirling approximation/ the derivative of $\zeta'(0)$). I want to avoid those. 
How I got an answer was via regularization of
$$-\sum_{i=1}^{\infty}\frac{\zeta(-i)}{i}=\sum_{n=1}^{\infty}\bigg(\left(n+\frac{1}{2}\right)\ln\left(1+ \frac{1}{n}\right)-1\bigg)$$
My own other try was rewriting it via: 
$$\sum_{n=1}^{\infty}\bigg(\left(n+\frac{1}{2}\right)\ln\left(1+\frac{1}{n}\right)-1\bigg)=\sum_{k=2}^{\infty} \zeta(k)(-1)^k \bigg(\frac{1}{k+1}-\frac{1}{2k}\bigg)$$
If this works I am already happy. If there's another simple way I'd love to hear it as well.

One has
\begin{align*}
& \sum\limits_{n = 1}^\infty  {\left[ {\left( {n + \frac{1}{2}} \right)\log \left( {1 + \frac{1}{n}} \right) - 1} \right]}  = \sum\limits_{n = 1}^\infty  {\int_0^1 {\frac{{\frac{1}{2} - t}}{{n + t}}dt} }  = \sum\limits_{n = 1}^\infty  {\int_0^1 {\frac{{\frac{1}{2} - (t - \left\lfloor t \right\rfloor )}}{{n + t}}dt} } \\ & = \sum\limits_{n = 1}^\infty  {\int_{n - 1}^n {\frac{{\frac{1}{2} - (t - \left\lfloor t \right\rfloor )}}{{t + 1}}dt} }  = \int_0^{ + \infty } {\frac{{\frac{1}{2} - (t - \left\lfloor t \right\rfloor )}}{{t + 1}}dt} .
\end{align*}
Now, by the Euler--Maclaurin formula,
$$
\log k! = \left( {k + \frac{1}{2}} \right)\log k- k + C + \int_0^{ + \infty } {\frac{{\frac{1}{2} - (t - \left\lfloor t \right\rfloor )}}{{t + k}}dt} 
$$
with some constant $C$. It can be shown that the integral is $\mathcal{O}(k^{-1})$ and so by Stirling's formula (or the Wallis product), $C=\frac{1}{2}\log (2\pi )$. Thus
\begin{align*}\sum\limits_{n = 1}^\infty  {\left[ {\left( {n + \frac{1}{2}} \right)\log \left( {1 + \frac{1}{n}} \right) - 1} \right]} & = \log 1! - \left( {\left( {1 + \frac{1}{2}} \right)\log 1 - 1 + \frac{1}{2}\log (2\pi )} \right) \\ &= 1 - \frac{1}{2}\log (2\pi ).
\end{align*}
