In their proof that $\log{n!}$ and $n\log{n}$ are asymptotically equivalent, Courant and Robbins derive the inequality $$L=1 - \frac{1}{\log{n}} + \frac{1}{n\log{n}}\lt \frac{\log{n!}}{n\log{n}} \lt \left(1+ \frac{1}{n}\right)\frac{\log{n}+\log\left(1+\frac{1}{n}\right) }{\log{n}}-\frac{1}{\log{n}}=U. $$
They then set the following exercise:
Prove that the two bounds are greater than $1-\frac{1}{n}$ and less than $1+ \frac{1}{n}$ respectively.
I interpreted this as asking us to prove $1-\frac{1}{n} \lt L$ and $ U \lt 1+\frac{1}{n}$, but, evaluating for specific (large) $n$, I found $1-\frac{1}{n} \gt L$, and thus I am unsure exactly what the question is asking us to prove?
Edit: Since @NN2 said the result holds asymptotically, I tried to evaluate the difference $$\left( 1-\frac{1}{n}\right) - L = \frac{1}{\log{n}} - \frac{1}{n\log{n}} - \frac{1}{n}.\tag{1}$$ By definition $\log{n} = x \iff e^x = n$. Substituting this into $(1)$, $$\frac{1}{x}-\frac{1}{xe^x}-\frac{1}{e^x}=\frac{e^x - (1+x)}{xe^x}\gt 0$$ for all $x\gt 0$, so again it seems like this isn’t the result the problem wants us to prove.