\(\rm \mathop {lim}\limits_{n \to \infty } \left( {\frac{{{{(n + 1)}^{1/3}}}}{{{n^{4/3}}}} + \frac{{{{(n + 2)}^{1/3}}}}{{{n^{4/3}}}} + \cdots \cdots + \frac{{{{(2n)}^{1/3}}}}{{{n^{4/3}}}}} \right)\) is equal to:

Concept:

\({\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to - \infty } \mathop \sum \limits_{{\rm{r}} = 1}^{{\rm{pn}}} \frac{1}{{\rm{n}}}{\rm{f}}\left( {\frac{{\rm{r}}}{{\rm{n}}}} \right) = \mathop \smallint \nolimits_0^{\rm{p}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}}} \)

Calculation:

The series given in the question is:

\(\rm \mathop {lim}\limits_{n \to \infty } \left( {\frac{{{{(n + 1)}^{1/3}}}}{{{n^{4/3}}}} + \frac{{{{(n + 2)}^{1/3}}}}{{{n^{4/3}}}} + \cdots \cdots + \frac{{{{(2n)}^{1/3}}}}{{{n^{4/3}}}}} \right)\)

\(\rm =\mathop {lim}\limits_{n \to \infty } \left( {\frac{{{{(n + 1)}^{1/3}}}}{{n.{n^{1/3}}}} + \frac{{{{(n + 2)}^{1/3}}}}{{n.{n^{1/3}}}} + \cdots \cdots + \frac{{{{(2n)}^{1/3}}}}{{n.{n^{1/3}}}}} \right)\)

Taking n^(1/3) outside and cancelling the same in the denominator, we get

\(\rm = \mathop {lim}\limits_{n \to \infty } \left( {\frac{{{{\left( {1 + \frac{1}{{\rm{n}}}} \right)}^{1/3}}}}{n} + \frac{{{{\left( {1 + \frac{2}{{\rm{n}}}} \right)}^{1/3}}}}{{\rm{n}}} + \cdots \cdots + \frac{{{{\left( {1 + \frac{{\rm{n}}}{{\rm{n}}}} \right)}^{1/3}}}}{{\rm{n}}}} \right)\)

\(\rm = \mathop {lim}\limits_{n \to \infty } \left( {{{\left( {1 + \frac{1}{{\rm{n}}}} \right)}^{1/3}} + {{\left( {1 + \frac{2}{{\rm{n}}}} \right)}^{1/3}} + \cdots \cdots + {{\left( {1 + \frac{{\rm{n}}}{{\rm{n}}}} \right)}^{1/3}}} \right)\frac{1}{n}\)

\(\rm = \mathop {lim}\limits_{n \to \infty } \mathop \sum \limits_{r = 1}^n {\left( {1 + \frac{r}{n}} \right)^{1/3}}\frac{1}{n}\)

Now taking, \(\rm \because \frac{r}{n}=x\text{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }\frac{1}{n}=\text{dx} \)

The equation becomes integral,

\(\rm = \mathop \smallint \limits_0^1 {\left( {1 + x} \right)^{\frac{1}{3}}}dx\)

\(\rm = \left[ {\frac{{{{\left( {1 + x} \right)}^{\frac{4}{3}}}}}{{\frac{4}{3}}}} \right]_0^1\)

\(\rm = \left[ {\frac{3}{4}{{(1 + x)}^{\frac{4}{3}}}} \right]_0^1\)

\(= \frac{3}{4}{\left( {1 + 1} \right)^{\frac{4}{3}}} - \frac{3}{4}{\left( {1 + 0} \right)^{\frac{4}{3}}}\)

\(= \frac{3}{4}{(2)^{\frac{4}{3}}} - \frac{3}{4}\)