Is it true that the square root always bounds this sum: $$\sqrt{n} \ge \limsup_{n \rightarrow \infty} \,\Biggr\lvert \sum_{m=1}^{n} \prod_{k \in \mathbb{P}}^{\infty} \Biggl(1-2 \biggl(\frac{\sin(\pi\,m)}{k \sin(\pi\frac{m}{k})}\biggr)^2\Biggr)\cdot\Biggl(1-\biggl(\frac{\sin(\pi\, m)}{k^2 \sin(\pi\frac{m}{k^2})} \biggr)^2\Biggr) \Biggl \lvert$$
I would really appreciate an answer, as I might use this for a paper I am writing (In the case that this is true). And I do hope that you can help me with this. $$$$ (If this is (not) true, could you please explain why)
Let $$\sigma(m,p)=\begin{cases}1&p\mid m\\0&p\nmid m\end{cases}$$ First note that \begin{align} \prod_{p \in \mathbb{P}}^{\infty} (1-2 \sigma(m,p))&=\prod_{p\mid m} (-1)\\ \prod_{p \in \mathbb{P}}^{\infty}(1-\sigma(m,p^2))&=|\mu(m)| \end{align} Consequently, $$\prod_{p \in \mathbb{P}}^{\infty} (1-2 \sigma(m,p))(1-\sigma(m,p^2))=\mu(m)$$ the Möbius function. Then $$\sum_{m=1}^{n} \prod_{p \in \mathbb{P}}^{\infty}(1-2 \sigma(m,p))(1-\sigma(m,p^2))=\sum_{m=1}^{n}\mu(m)=M(n)$$ is the Mertens function. So you are asking $|M(n)|\leq\sqrt n$ as $n\to+\infty$,
which is not known today, that's the Mertens Conjecture.