$(2016+1)+(2017+1) = (4034+1)$

This is just a repeated application of a common technique used to evaluate limits involving series. The idea is to first rewrite the limit with summation notation and then try to spot what Riemann sum is similar to the series. First, rewrite everything as $\frac{\sum_{k=1}^n k^{2016} \sum_{k=1}^n k^{2017}}{\sum_{k=1}^n k^{4034}}$ . Using a little algebra this is equal to $\frac{n^{2016} \times n \times \frac{1}{n} \sum_{k=1}^n \left (\frac{k}{n} \right)^{2016} n^{2017} \times n \times \frac{1}{n}\sum_{k=1}^n \left (\frac{k}{n} \right)^{2017}}{n^{4034} \times n \times \frac{1}{n}\sum_{k=1}^n\left (\frac{k}{n} \right)^{4034}} = \frac{ \frac{1}{n} \sum_{k=1}^n \left (\frac{k}{n} \right)^{2016} \frac{1}{n}\sum_{k=1}^n \left (\frac{k}{n} \right)^{2017}}{ \frac{1}{n}\sum_{k=1}^n\left (\frac{k}{n} \right)^{4034}}$ . These are now the Riemann sums for the functions $x^{2016}, x^{2017}, x^{4034}$ over the interval $(0,1)$ so if we take the limit as $n \to \infty$ then the Riemann sums become the integral so the limit is equal to $\frac{\int_{0}^1 x^{2016} dx \int_{0}^1 x^{2017} dx}{\int_{0}^1 x^{4034} dx}$ and by applying the fundamental theorem of calculus this evaluates to $\frac{\frac{1}{2017} \times \frac{1}{2018}}{\frac{1}{4035}}$ .