Find $\lim_{n \rightarrow \infty}\frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}$ Find:
$$\lim_{n \rightarrow \infty} \frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}$$
The sequence $\frac{1}{nx^2 \log{(1+ \frac{x}{n})}}=\frac{1}{x^3 \frac{\log{(1+ \frac{x}{n})}}{\frac{x}{n}}}$ converges pointwise to $\frac{1}{x^3}$. So if we could apply Lebesgue's Dominated Convergence Theorem, we have:
$\lim_{n \rightarrow \infty} \frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}=\lim_{n \rightarrow \infty} \int_{1}^{\infty} \frac{\mathrm dx}{x^3}=\frac{1}{2}$
I have a problem with finding a majorant. Could someone give me a hint?
 A: I showed in THIS ANSWER, using only Bernoulli's Inequality the sequence $\left(1+\frac xn\right)^n$ is monotonically increasing for $x>-n$.  
Then, we can see that for $x\ge 1$ and $n\ge1$, the sequence $f_n(x)$ given by
$$f_n(x)=n\log\left(1+\frac xn\right)$$
is also monotonically increasing.  Therefore, a suitable dominating function is provided simply by the inequality
$$\frac{1}{n\log\left(1+\frac xn\right)}\le \frac{1}{\log(1+x)}\le \frac{1}{\log(2)}$$
Therefore, we have
$$\frac{1}{nx^2\log\left(1+\frac xn\right)}\le \frac{1}{x^2\log(2)}$$
Using the dominated convergence theorem, we can assert that 
$$\begin{align}
\lim_{n\to \infty}\int_1^\infty \frac{1}{nx^2\log\left(1+\frac xn\right)}\, dx&=\int_1^\infty \lim_{n\to \infty}\left(\frac{1}{nx^2\log\left(1+\frac xn\right)}\right)\,dx\\\\
&=\int_1^\infty\frac{1}{x^3}\,dx\\\\
&=\frac12.
\end{align}$$
A: I think one could do this in a conceptually simpler way:
Since
$$
\frac{y}{1+y}<\log(1+y)<y
$$
your integrand is bounded as
$$
\frac{1}{x^3}<\frac{1}{nx^2\ln(1+x/n)}<\frac{1}{nx^2(x/n)/(1+x/n)}=\frac{1}{x^3}(1+x/n).
$$
By monotonicity, your integral satisfies
$$
\frac{1}{2}=\int_1^{+\infty}\frac{1}{x^3}\,\mathrm dx<\int_1^{+\infty}\frac{1}{nx^2\ln(1+x/n)}\,\mathrm dx<\int_1^{+\infty}\frac{1}{x^3}(1+x/n)\,\mathrm dx=\frac{1}{2}+\frac{1}{n}.
$$
Now squeeze.
