_ n [ n^2 n^3 ] = - Vidyadip

MCQ

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\Sigma {n^2}}}{{{n^3}}}} \right] = $

A
$ - \frac{1}{6}$
B
$\frac{1}{6}$
✓
$\frac{1}{3}$
D
$ - \frac{1}{3}$

✓

Answer

Correct option: C.

$\frac{1}{3}$

c
(c) $\mathop {\lim }\limits_{n \to \infty } \,\left[ {\frac{{n\,(n + 1)\,(2n + 1)}}{{6{n^3}}}} \right] = \mathop {\lim }\limits_{n \to \infty } \,\frac{{\left( {1 + \frac{1}{n}} \right)\,\left( {2 + \frac{1}{n}} \right)}}{6} = \frac{1}{3}.$

Note : Students should remember that

$\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum n}}{{{n^2}}} = \frac{1}{2},\,\,\,\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum {n^2}}}{{{n^3}}} = \frac{1}{3}$

and $\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum {n^3}}}{{{n^4}}} = \frac{1}{4}.$

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