Gentle sum 2-riemann integral

The above series sums to $\frac{1}{n^4} \cdot \frac{n^{2}(n+1)^{2}}{4} = \frac{n^4 + 2n^3 + n^2}{4n^4} = \frac{1}{4} + \frac{1}{2n} + \frac{1}{4n^2}.$ The final limit as $n \rightarrow \infty$ is $\boxed{\frac{1}{4}}.$