Find $\lim_{n \rightarrow \infty}\frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}$

Question

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?

Answer 1

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.

Answer 2

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

$$\begin{align} \lim_{n\to \infty}\int_1^\infty \frac{1}{nx^2\log\left(1+\frac xn\right)}\, dx&=\int_1^\infty \lim_{n\to \infty}\left(\frac{1}{nx^2\log\left(1+\frac xn\right)}\right)\,dx\\\\ &=\int_1^\infty\frac{1}{x^3}\,dx\\\\ &=\frac12. \end{align}$$