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(B) Let $S = \mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^n \frac{r^2}{r^3 + n^3}$.We can rewrite the sum as:$S = \mathop {\lim }\limits

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$\frac{1}{3}{\log _e}2$

Vedclass · Answer

(B) Let $S = \mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^n \frac{r^2}{r^3 + n^3}$.We can rewrite the sum as:$S = \mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^n \frac{r^2}{n^3 \left( \left( \frac{r}{n} ight)^3 + 1 ight)}$.Multiplying and dividing by $n$,we get:$S = \mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^n \frac{1}{n} \cdot \frac{(r/n)^2}{1 + (r/n)^3}$.Using the definition of a definite integral as the limit of a sum,$\mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^n \frac{1}{n} f\left( \frac{r}{n} ight) = \int_0^1 f(x) dx$,where $f(x) = \frac{x^2}{1 + x^3}$.Thus,$S = \int_0^1 \frac{x^2}{1 + x^3} dx$.Let $u = 1 + x^3$,then $du = 3x^2 dx$,or $x^2 dx = \frac{1}{3} du$.When $x = 0, u = 1$. When $x = 1, u = 2$.$S = \frac{1}{3} \int_1^2 \frac{1}{u} du = \frac{1}{3} [\log_e u]_1^2 = \frac{1}{3} (\log_e 2 - \log_e 1) = \frac{1}{3} \log_e 2$.

$\mathop {\lim }\limits_{n \to \infty } \left( \frac{1^2}{1^3 + n^3} + \frac{2^2}{2^3 + n^3} + \dots + \frac{n^2}{n^3 + n^3} \right)$ is equal to

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