=\lim_{n\to \infty } \frac{n\left( \ln \left(\frac{n+1}{n}\right)+1\right)}{2(n+1)}-\lim_{n\to \infty } \frac{\frac{1}{2}\ln (2 \pi  n)}{2(n+1)}=\lim_{n\to \infty } \frac{n}{2(n+1)}\lim_{n\to \infty } \left( \ln \left(\frac{n+1}{n}\right)+1\right)-\frac{1}{4}\lim_{n\to \infty } \frac{\ln (2 \pi )+\ln (n)}{n+1}=\frac{1}{2}