_n [(n+1)^1/3n^4/3 + (n+2)^1/3n^4/3 + + (2n)^1/3n^4/3] is equal to:

Question

$\displaystyle\lim_{n	o\infty} \left[\frac{(n+1)^{1/3}}{n^{4/3}} + \frac{(n+2)^{1/3}}{n^{4/3}} + \cdots + \frac{(2n)^{1/3}}{n^{4/3}}ight]$ is equal to:

Accepted Answer

$$\mathop {\lim }\limits_{n \to \infty } {{{{(n + 1)}^{{1 \over 3}}} + {{(n + 2)}^{{1 \over 3}}} + ..... + {{\left( {n + n} \right)}^{{1 \over 3}}}} \over {n{{(n)}^{{1 \over 3}}}}}$$

$$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{{{(n + r)}^{{1 \over 3}}}} \over {n.{n^{{1 \over 3}}}}}\,\,\,{r \over n} \to x\,and\,{1 \over n} \to dx\,} $$

$$ \Rightarrow \int\limits_0^1 {{{(1 + x)}^{{1 \over 3}}}dx} $$

$$ \Rightarrow \left[ {{3 \over 4}{{(1 + x)}^{{4 \over 3}}}} \right]_0^1 = {3 \over 4}{(2)^{{4 \over 3}}} - {3 \over 4}$$

Solutions · 1