Study the convergence of the following series as $\alpha>0$
$$\sum_{n=1}^{+\infty}\int_{n^{\alpha}}^{n^{\alpha+1}}\log^2\left(1+\sin\frac{1}{x}\right)\,dx$$
This is the first time I do an exercise with a series defined by a integral and I hold it is equal to study the convergence of the improper integral (isn't it?):
$$\lim_{N \to +\infty}\int_{1^{\alpha}}^{N^{\alpha+1}}\log^2\left(1+\sin\frac{1}{x}\right)\,dx$$
or, do I need to follow a different path?
Note that for small $\epsilon$, $$\log^2 (1+\epsilon) \approx \epsilon^2$$ and $$\sin \epsilon \approx \epsilon$$ so for large $x$, $$\log^2 \left( 1 + \sin \frac{1}{x} \right) \approx \frac{1}{x^2}.$$ Therefore, $$\int_{n^\alpha}^{n^{\alpha+1}} \log^2 \left( 1 + \sin \frac{1}{x} \right)\, dx \approx \int_{n^\alpha}^{n^{\alpha+1}} \frac{dx}{x^2} = \frac{1}{n^\alpha} - \frac{1}{n^{\alpha+1}}.$$ Note that $\sum_{n=1}^\infty \frac{1}{n^\alpha}$ converges iff $\alpha > 1$, but $\sum_{n=1}^\infty \frac{1}{n^{\alpha+1}}$ always converges, so we have convergence iff $\alpha > 1$.