JEE MainMathematicsEasyLimits

$\lim_{n \rightarrow \infty}\mspace{2mu}\frac{1^{2} + 2^{2} + 3^{2} + \ldots\ldots\ldots + n^{2}}{n^{3}} =$

Show answer and solution

\lim_{n \rightarrow \infty}\mspace{2mu}\frac{1^{2} + 2^{2} + 3^{2} + \ldots\ldots\ldots + n^{2}}{n^{3}} = \lim_{n \rightarrow \infty}\mspace{2mu}\frac{n(n + 1)(2n + 1)}{6n^{3}}

= \frac{1}{6}\lim_{n \rightarrow \infty}\mspace{2mu}\frac{1\left( 1 + \frac{1}{n} \right)\left( 2 + \frac{1}{n} \right)}{1} = \frac{1}{6}(1 + 0)(2 + 0) = \frac{1}{3}

Source: Grass-G / M-19 Mathematics Paper 28 Nov Fourth.docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.