Question

$\underset{n\to \infty }{\lim }\left(\frac{1^2+2^2+\dots ..+n^2}{n^3}\right)=$

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{1}{6}$

Answer 1

The correct option is C

$\frac{1}{3}$

Explanation for the correct option:

Finding the value of the given limit:

$$\begin{gathered} \underset{n\to \infty }{\lim }\left(\frac{1^2+2^2+\dots ..+n^2}{n^3}\right) \\ =\underset{n\to \infty }{\lim }\left(\frac{n(n+1)(2n+1)}{6n^3}\right)\left[\because \sum _{}{n}^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right] \\ =\underset{n\to \infty }{\frac{1}{6}\lim }\left(\frac{(n+1)(2n+1)}{n^2}\right) \\ =\frac{1}{6}\underset{n\to \infty }{\lim }\left(\frac{2n^2+n+2n+1}{n^2}\right) \\ =\frac{1}{6}\underset{n\to \infty }{\lim }\left(\frac{2n^2+3n+1}{n^2}\right) \\ =\frac{1}{6}\underset{n\to \infty }{\lim }\left(2+\frac{3}{n}+\frac{1}{n^2}\right) \end{gathered}$$

Applying the limits,

$$\begin{gathered} =\frac{1}{6}\left(2+0+0\right)\left[\because \frac{x}{\infty }=0\right] \\ =\frac{2}{6} \\ =\frac{1}{3} \end{gathered}$$

Therefore, the correct answer is option (C).