limn→∞ 12+22+32+…+n2n3= - Infinity Learn by Sri Chaitanya

A
$\frac{1}{6}$
B
$\frac{1}{3}$
C
$\frac{1}{2}$
D
none of these

Solution:

we have,

$\begin{array}{l} lim_{n \to \infty} \frac{1^{2} + 2^{2} + 3^{2} + \dots + n^{2}}{n^{3}} \\ = lim_{n \to \infty} \frac{n (n + 1) (2 n + 1)}{6 n^{3}} = \frac{2}{6} = \frac{1}{3} \end{array}$

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