## Calculus (3rd Edition)

$b_n$ converges to $1$.
We have $$\lim _{n \rightarrow \infty} b_n=\lim _{n\rightarrow \infty} n(\ln(n+1)-\ln n) \\ =\lim _{n\rightarrow \infty} n\ln\frac{n+1}{n}=\lim _{n\rightarrow \infty} n\ln\frac{n+1}{n}\\ =\lim _{n\rightarrow \infty} \ln\left(1+\frac{1}{n}\right)^n= \ln\lim _{n\rightarrow \infty} \left(1+\frac{1}{n}\right)^n=\ln e=1$$ Where we used the fact that $\lim _{m \rightarrow \infty} \left(1+\frac{1}{m}\right)^{m}=e$. So, $b_n$ converges to $1$.