Determine the convergence of the improper integral

Question

This is link to question [Here is my attempt, but the answer key is convergent. I dont think I count it wrong.][1]

Answer 1

Since the integral $\int_1^\infty\frac{\mathrm dx}{1+\sqrt x}$ diverges, you cannot deduce from the fact that $(\forall x\in[1,\infty)):\frac{\sin(1/x)}{2+\sqrt x}\leqslant\frac1{2+\sqrt x}$ that your integral converges.

Answer 2

$f(x) = \frac{\sin(1/x)}{2+\sqrt{x}} \sim \frac{1/x}{2+\sqrt{x}} \sim \frac{1}{x^{3/2}}$ as $x\to +\infty$.

Moreover, $f(x) \geq 0 $ for $x$ sufficiently large.

Therefore, $\int_1^{+\infty} f(x) dx $ converges since $\int_1^{+\infty}\frac{1}{x^{3/2}} dx$ converges.