Aloha :)
$$a_n=\frac{(\sqrt{n^5}+2n)\left(\sin^2\left(\frac{1}{n}\right)+1\right)}{(n+1)^2\sqrt[3]{1+2n}}>\frac{(\sqrt{n^5}+2n)}{(n+1)^2\sqrt[3]{1+2n}}>\frac{\sqrt{n^5}}{(n+1)^2\sqrt[3]{1+2n}}$$$$\phantom{a_n}=\frac{n^2\cdot n^{1/2}}{(n+1)^2(1+2n)^{1/3}}=\frac{n^2}{(n+1)^2}\,\frac{n^{3/6}}{(1+2n)^{2/6}}=\frac{n^2}{(n+1)^2}\,\left(\frac{n^3}{(1+2n)^2}\right)^{1/6}$$$$\phantom{a_n}=\frac{n^2}{(n+1)^2}\,\left(\frac{n^3}{(1+2n)^2}\right)^{1/6}=\frac{n^2}{(n+1)^2}\,\left(\frac{n^3}{1+4n+4n^2}\right)^{1/6}$$$$\phantom{a_n}=\frac{n^2}{(n+1)^2}\,\left(\frac{n^2}{1+4n+4n^2}\right)^{1/6}\!\!\!\sqrt[6]{n}=\underbrace{\frac{1}{(1+\frac{1}{n})^2}}_{\to1}\,\underbrace{\left(\frac{1}{\frac{1}{n^2}+\frac{4}{n}+4}\right)^{1/6}}_{\to2^{-1/3}}\!\!\left(\underbrace{\sqrt[6]{n}}_{\to\infty}\right)\to\infty$$
