\lim_{n\to \infty } \frac{\ln (2 \pi )+\ln (n)}{n+1}=\lim_{n\to \infty } \frac{\ln (2 \pi )}{n+1}+\lim_{n\to \infty } \frac{\ln (n)}{n+1}=0+\lim_{n\to \infty } \ln \left(n^{\frac{1}{n+1}}\right)=\lim_{n\to \infty } \ln \left(n^{\frac{1}{n}}\right)^{\frac{n}{n+1}}=\ln (1)^{\lim_{n\to \infty } \frac{n}{n+1}}=\ln  1=0