Evaluate the following limit: _ n 1^2+2^2+ +n^2 n^3 - Vidyadip

Question

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

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Answer

$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^2+2^2+\ \cdots+\text{n}^2}{\text{n}^3}$ $=\lim\limits_{\text{n}\rightarrow\infty}\frac16\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{\text{n}^3}$ $\Big[1^2+2^2+3^2+\ \cdots+\text{n}^2=\frac{1}{6}\text{n}(\text{n}+1)(2\text{n}+1)\Big]$ $=\frac{1}{6}\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(\text{n}^2+\text{n}\big)(2\text{n}+1)}{\text{n}^3}$ $=\frac{1}{6}\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(2\text{n}^3+\text{n}^2+2\text{n}^2+\text{n}\big)}{\text{n}^3}$ $=\frac16\lim\limits_{\text{n}\rightarrow{\infty}}\frac{\big(2\text{n}^2+3\text{n}^2+\text{n}\big)}{\text{n}^3}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$ $=\frac16\lim\limits_{\text{n}\rightarrow{\infty}}\frac{\Big(2+\frac{3}{\text{n}}+\frac{1}{\text{n}^2}\Big)}{1}$ $=\frac16\frac{(2+0+0)}{1}=\frac{1}{3}$

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