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Question: \(\lim _ { n \rightarrow \infty }\) \(\left\lbrack \frac{1}{n} + \frac{n^{2}}{(n + 1)^{3}} + \frac{n...

lim⁡n→∞\lim _ { n \rightarrow \infty } [1n+n2(n+1)3+n2(n+2)3+...+18n]\left\lbrack \frac{1}{n} + \frac{n^{2}}{(n + 1)^{3}} + \frac{n^{2}}{(n + 2)^{3}} + ... + \frac{1}{8n} \right\rbrack is equal to –

A

38\frac{3}{8}

B

14\frac{1}{4}

C

18\frac{1}{8}

D

None of these

Answer

38\frac{3}{8}

Explanation

Solution

lim⁡n→∞\lim_{n \rightarrow \infty}

= [n2(n+0)3+n2(n+1)3+n2(n+2)3+...+n2(n+n)3]\left\lbrack \frac{n^{2}}{(n + 0)^{3}} + \frac{n^{2}}{(n + 1)^{3}} + \frac{n^{2}}{(n + 2)^{3}} + ... + \frac{n^{2}}{(n + n)^{3}} \right\rbrack

=lim⁡n→∞\lim_{n \rightarrow \infty} ∑r=0nn2(n+r)3\sum_{r = 0}^{n}\frac{n^{2}}{(n + r)^{3}}= lim⁡n→∞\lim _ { n \rightarrow \infty } ∑r=0n1n\sum_{r = 0}^{n}\frac{1}{n}.1(1+rn)3\frac{1}{\left( 1 + \frac{r}{n} \right)^{3}}

= ∫01dx(1+x3)\int_{0}^{1}\frac{dx}{(1 + x^{3})}= [−12(1+x)2]01\left\lbrack - \frac{1}{2(1 + x)^{2}} \right\rbrack_{0}^{1}= –12\frac { 1 } { 2 } (14−1)\left( \frac{1}{4} - 1 \right) = 38\frac{3}{8}.

Hence (1) is the correct answer.