I want to know if my reasoning is correct about the divergence of the series $\displaystyle \sum_{n=1}^{\infty} \frac{\cos (\frac{1}{n^2})}{\sqrt{2n+1}}$.
First, we know that $\displaystyle \frac{1}{n^2} \in (0,1], \forall n \in \mathbb{N}$. The function $\cos x$ is decreasing and positive on the interval $(0,1]$ and so $\cos 1 >0$.
Also, we have $\displaystyle 1 \geq \frac{1}{n^2}, \forall n \in \mathbb{N} \implies \cos 1 \leq \cos(\frac{1}{n^2}), \forall n \in \mathbb{N}$.
So, $\displaystyle a_n = \frac{\cos (\frac{1}{n^2})}{\sqrt{2n+1}} \geq \frac{\cos 1}{\sqrt{2n+1}} = b_n$. Because, $\displaystyle \sum_{n=1}^{\infty}b_n$ diverges, then by the comparison test $\displaystyle \sum_{n=1}^{\infty}a_n$ diverges too.
Thanks for any advices!