limn→∞ 1n+n2(n+1)3+n2(n+2)3+…+18n

limn→∞ 1n+n2(n+1)3+n2(n+2)3+…+18n=limn→∞ n2(n+0)3+n2(n+1)3+n2(n+2)3+…+n2(n+n)3=limn→∞ ∑r=0n n2(n+r)3=limn→∞ ∑r=0n n2(n(1+rn)3=limn→∞ 1n∑r=0n 1(1+(r/n))3=∫01 dx(1+x)3=−12(x+1)201=12−18=38

limn→∞ 1n+n2(n+1)3+n2(n+2)3+…+18n

limn→∞ 1n+n2(n+1)3+n2(n+2)3+…+18n=limn→∞ n2(n+0)3+n2(n+1)3+n2(n+2)3+…+n2(n+n)3=limn→∞ ∑r=0n n2(n+r)3=limn→∞ ∑r=0n n2(n(1+rn)3=limn→∞ 1n∑r=0n 1(1+(r/n))3=∫01 dx(1+x)3=−12(x+1)201=12−18=38

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$lim_{n \to \infty} (\frac{1}{n} + \frac{n^{2}}{(n + 1)^{3}} + \frac{n^{2}}{(n + 2)^{3}} + \dots + \frac{1}{8 n})$

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$\frac{3}{2}$

$\frac{3}{8}$

$\frac{2}{8}$

$\frac{2}{3}$

answer is D.

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Detailed Solution

$\begin{array}{l} lim_{n \to \infty} (\frac{1}{n} + \frac{n^{2}}{(n + 1)^{3}} + \frac{n^{2}}{(n + 2)^{3}} + \dots + \frac{1}{8 n}) \\ = lim_{n \to \infty} (\frac{n^{2}}{(n + 0)^{3}} + \frac{n^{2}}{(n + 1)^{3}} + \frac{n^{2}}{(n + 2)^{3}} + \dots + \frac{n^{2}}{(n + n)^{3}}) \\ = lim_{n \to \infty} \sum_{r = 0}^{n} (\frac{n^{2}}{(n + r)^{3}}) \\ = lim_{n \to \infty} [\sum_{r = 0}^{n} (\frac{n^{2}}{(n (1 + \frac{r}{n})^{3}})] \\ = lim_{n \to \infty} [\frac{1}{n} \sum_{r = 0}^{n} (\frac{1}{(1 + (r / n))^{3}})] \\ \begin{aligned} = \int_{0}^{1} (\frac{d x}{(1 + x)^{3}}) \\ = {(- \frac{1}{2 (x + 1)^{2}})}_{0}^{1} \\ = |\frac{1}{2} - \frac{1}{8}| = \frac{3}{8} \end{aligned} \end{array}$