Check for Convergence

$\begin{aligned} S & = \sum_{n=1}^\infty \sqrt{\frac {n-n\sin \frac 1n}{n^4}} \\ & = \sum_{n=1}^\infty \sqrt{\frac {1-{\color{#3D99F6}\sin \frac 1n}}{n^3}} & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \sum_{n=1}^\infty \sqrt{\frac {1-{\color{#3D99F6}\left( \frac 1n - \frac 1{3!n^3}+\frac 1{5!n^5} - ... \right)}}{n^3}} \\ & = \sum_{n=1}^\infty \sqrt{\frac 1{n^3} - \frac 1{n^4} + \frac 1{3!n^6} + \frac 1{5!n^8} - ... } \\ & < \sum_{n=1}^\infty \sqrt{\frac 1{n^3}} = {\color{#3D99F6}\zeta \left(\frac 32\right)} \approx 2.612 & \small \color{#3D99F6} \zeta(\cdot) \text{ is the Riemann zeta function} \end{aligned}$

Therefore, the sum $\displaystyle S = \sum_{n=1}^\infty \sqrt{\frac {n-n\sin \frac 1n}{n^4}}$ converges to a finite value.

By the Weierstrass test or the M-test the given series is less than $\large\frac{1}{n^2}$ which is a $p-$ series of degree $2$ which converges. Technically, it is similar to the comparison test of series, in some way.