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}$
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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$15$ |
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$10$ |