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Question

Question: \[\lim_{n \rightarrow \infty}\left\lbrack \frac{\sum n^{2}}{n^{3}} \right\rbrack =\]...

lim⁡n→∞[∑n2n3]=\lim_{n \rightarrow \infty}\left\lbrack \frac{\sum n^{2}}{n^{3}} \right\rbrack =

A

−16- \frac{1}{6}

B

16\frac{1}{6}

C

13\frac{1}{3}

D

−13\frac{- 1}{3}

Answer

13\frac{1}{3}

Explanation

Solution

lim⁡n→∞[n(n+1)(2n+1)6n3]=lim⁡n→∞(1+1n)(2+1n)6=13\lim_{n \rightarrow \infty}\left\lbrack \frac{n(n + 1)(2n + 1)}{6n^{3}} \right\rbrack = \lim_{n \rightarrow \infty}\frac{\left( 1 + \frac{1}{n} \right)\left( 2 + \frac{1}{n} \right)}{6} = \frac{1}{3}

Note : Students should remember that,

lim⁡n→∞∑nn2=12andlim⁡n→∞∑n2n3=13.\lim_{n \rightarrow \infty}\frac{\sum n}{n^{2}} = \frac{1}{2}\text{and}\lim_{n \rightarrow \infty}\frac{\sum n^{2}}{n^{3}} = \frac{1}{3}.