An approximation of $\log \frac{n+1}{n}$ inside of a limit This question arises from a question posed here:
$$\lim_{n \to \infty} n - n^2 \log \frac{n+1}{n} = \frac{1}{2}$$
I figured we could approximate $\log \frac{n+1}{n} = \int_{n}^{n+1} \frac{1}{x} \, dx \approx \frac{1}{2} ( \frac{1}{n} + \frac{1}{n+1} )$ and we get the right answer if we plug that in and then take the limit. This method is worrying, however, since we seem to have applied limits separately.
Indeed, if we take any different approximation of the $\log$ term of the form $\frac{\alpha}{n} + \frac{(1 - \alpha)}{n+1}$ with $\alpha \in (0,1)$ we get the limit to be $1 - \alpha$. I would actually think some $\alpha < 1/2$ would be ideal since $\log$ is concave, but I suppose as $n \to \infty$ the behavior of log on any interval $[n, n+1]$ becomes more and more linear, hence why we take $\alpha = \frac{1}{2}$.
This is a lot of text; I suppose the question is: is it rigorous to approximate $\log$ like this and then apply the limit separately to the rest of the term, or are we just lucky that it works out? If it is not rigorous, what arguments can we make to better understand why and convince ourselves that it works out?
 A: You should check the asymptotics of this approximation, i.e. the precision of the $\approx$ sign in front of $\frac{1}{2}$. I don't know how sharp this is, therefore I'd rather use Taylor series for $\log (1-t) \sim -t +\frac{t^2}{2}$. 
A: On the one hand, as $n \to \infty$, you have
$$
\frac{1}{2} \left( \frac{1}{n} + \frac{1}{n+1} \right)=\frac{1}{2n} \left(1+ \frac{1}{1+\frac1n}\right)=\frac{1}{2n} \left( 2- \frac{1}{n}+O\left(\frac{1}{n^2}  \right)\right)=\color{red}{\frac{1}{n}+O\left(\frac{1}{n^2}  \right)}.
$$ On the other hand, by the Taylor expansion,
$$
\log \frac{n+1}{n}=\log\left(1 + \frac{1}{n}\right) = \color{red}{\frac{1}{n}+O\left(\frac{1}{n^2}  \right)}.
$$  Thus the approximation is valid up to $\displaystyle O\left(\frac{1}{n^2}  \right)$.
A: The integral of the function $f(x)$ over the interval $[a,b]$ can be written using the Trapezoidal Rule with correction term
$$\int_a^b f(x)\,dx=\left(\frac{f(a)+f(b)}{2}\right)(b-a)-\frac1{12}f''(\xi)(b-a)^3$$
for some $\xi\in[a,b]$.  
Here, we have $f(x)=1/x$ and $f''(x)=2/x^3$.   Thus, we can write
$$\begin{align}
\log\left(\frac{n+1}{n}\right)&=\int_n^{n+1}\frac1x\,dx\\\\
&=\frac12\left(\frac1{n+1}+\frac1n\right)-\frac16\frac{1}{\xi^3}
\end{align}$$
Finally, we assert that 
$$ -\frac{1}{6n^3}<\log\left(\frac{n+1}{n}\right)-\frac12\left(\frac1{n+1}+\frac1n\right)<-\frac{1}{6(n+1)^3}$$
for some $n<\xi<n+1$.
