Question
Evaluate the following limit:
Evaluate: $\lim\limits_{\text{n}\rightarrow\infty}\frac{1^4+2^4+3^4+\ \cdots+\text{n}^4}{\text{n}^5}-\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+\ \cdots+\text{n}^3}{\text{n}^5}$

Answer

$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^4+2^4+3^4+\ \cdots+\text{n}^4}{\text{n}^5}-\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+\ \cdots+\text{n}^3}{\text{n}^5}$
$=\ \lim\limits_{\text{n}\rightarrow\infty}\frac{\frac{\text{n}(1+\text{n})(1+2\text{n})\big(-1+3\text{n}+3\text{n}^2\big)}{30}}{\text{n}^5}-\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)^2}{\text{n}^5}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(\frac{1}{\text{n}}+1\big)\big(\frac{1}{\text{n}}+2\big)\big(-\frac{1}{\text{n}^2}+\frac{3}{\text{n}}+3\big)}{30}\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}^5}\Bigg(\frac{\text{n}^2\big(\text{n}^2+2\text{n}+1\big)}{4}\Bigg)$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(\frac{1}{\text{n}}+1\big)\big(\frac{1}{\text{n}}+2\big)\big(-\frac{1}{\text{n}^2}+\frac{3}{\text{n}}+3\big)}{30}\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big(\frac{1}{\text{n}}+\frac{2}{\text{n}^2}+\frac{1}{\text{n}^3}\Big)}{4}$
$=\frac{1\times2\times3}{30}-0$
$=\frac15$

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