Limit of sequence using integral test and L'Hopital's rule

$$\lim_{n \to \infty}\frac{1}{\log n}\sum_{k=1}^{n^2}1/k$$

My attempt:$$\lim_{n \to \infty}\sum_{k=1}^{n^2}1/k=\lim_{n \to\infty} \int_1^{n^2}1/xdx$$

Then this converges to infinity as it is harmonic series, and $$\lim_{n \to \infty}\log n \to \infty $$Using L'hopital Rule,$$\lim_{n \to \infty}\frac{1}{\log n}\sum_{k=1}^{n^2}1/k=\lim_{n \to\infty}\frac{1/n^2-1}{1/n}=1$$

My question: Is this correct? If so is it possible everytime to make such series of decreasing sequence to integration ( integral test) whose endpoints are this way.If not correct please help me.