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$\large\color{#EC7300}{\psi}=\dfrac{\displaystyle\lim_{n\rightarrow\infty}\left(\dfrac{1^{2016}+2^{2016}+\cdots+n^{2016}}{n^{2017}}\right)}{\displaystyle\lim_{n\rightarrow\infty}\left(\dfrac{1^{2017}+2^{2017}+\cdots+n^{2017}}{n^{2018}}\right)}$ $\large =\dfrac{\displaystyle\lim_{n\rightarrow\infty}\dfrac 1n \displaystyle\sum_{r=1}^{n}\left(\dfrac rn\right)^{2016}}{\displaystyle\lim_{n\rightarrow\infty}\dfrac 1n \displaystyle\sum_{r=1}^{n}\left(\dfrac rn\right)^{2017}}$ Using Riemann Sums $\Large =\dfrac{\int_0^1 x^{2016}}{\int_0^1 x^{2017}}$ $\large =\dfrac{2018}{2017}=1+\dfrac{1}{2017}$ $\large\color{#302B94}{\beta=2017}$ $2(2017-1)=63(64)\implies \color{forestgreen}{\phi=63}$ $\Large\mathfrak{P}=\sum_{n=\color{forestgreen}{4}}^{\color{forestgreen}{63}}\left(\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)$ $\large\color{#69047E}{\textbf{(A Telescopic Series)}}$ $\large=\dfrac 15-\dfrac{1}{65}=\dfrac{12}{65}=\color{#D61F06}{\dfrac{\delta}{\alpha}}$ $\therefore \Huge \dfrac{65}{13}= \Huge\color{#456461}{\boxed{\color{#D61F06}{\boxed{\Huge\color{#007fff}{5}}}}}$