Question

Find the value of following limit:
$\displaystyle \lim_{n\to \infty }\left( \dfrac{{{\left( n+1 \right)}^{\dfrac{1}{3}}}}{{{n}^{\dfrac{4}{3}}}}+\dfrac{{{\left( n+2 \right)}^{\dfrac{1}{3}}}}{{{n}^{\dfrac{4}{3}}}}+...........+\dfrac{{{\left( 2n \right)}^{\dfrac{1}{3}}}}{{{n}^{\dfrac{4}{3}}}} \right)$
(A) $\dfrac{4}{3}{{\left( 2 \right)}^{\dfrac{4}{3}}}$
(B) $\dfrac{3}{2}{{\left( 2 \right)}^{\dfrac{4}{3}}}-\dfrac{4}{3}$
(C) $\dfrac{3}{4}{{\left( 2 \right)}^{\dfrac{4}{3}}}-\dfrac{3}{4}$
(D) $\dfrac{4}{3}{{\left( 2 \right)}^{\dfrac{3}{4}}}$

Answer 1

Here for solving this question, we will simplify the given equation by making some required simplifications into the form of $\int_{a}^{b}{f(x)dx}=\displaystyle \lim_{n\to \infty }h\sum\limits_{r=0}^{n}{f(a+rh)}$ where $h=\dfrac{b-a}{n}$ $$$$ this equation and will get our required option.

Complete step by step answer:
In this question we need to find the value of:
$\displaystyle \lim_{n\to \infty }\left( \dfrac{{{\left( n+1 \right)}^{\dfrac{1}{3}}}}{{{n}^{\dfrac{4}{3}}}}+\dfrac{{{\left( n+2 \right)}^{\dfrac{1}{3}}}}{{{n}^{\dfrac{4}{3}}}}+...........+\dfrac{{{\left( 2n \right)}^{\dfrac{1}{3}}}}{{{n}^{\dfrac{4}{3}}}} \right)$
By applying the concept of summation, this equation can be simply written as
$\begin{aligned} & \displaystyle \lim_{n\to \infty }\dfrac{1}{{{n}^{\dfrac{4}{3}}}}\left( \sum\limits_{r=0}^{r=n}{{{\left( n+r \right)}^{\dfrac{1}{3}}}} \right) \\\ & \Rightarrow \displaystyle \lim_{n\to \infty }\dfrac{1}{n}\left( {{\sum\limits_{r=0}^{r=n}{\left( \dfrac{n+r}{n} \right)}}^{\dfrac{1}{3}}} \right) \\\ \end{aligned}$
By taking n common in the denominator
$\Rightarrow \displaystyle \lim_{n\to \infty }\dfrac{1}{n}\left( {{\sum\limits_{r=0}^{r=n}{\left( 1+\dfrac{r}{n} \right)}}^{\dfrac{1}{3}}} \right)$
As we know that from the concept of integration for the summation of continuous part, $\int_{a}^{b}{f(x)dx}=\displaystyle \lim_{n\to \infty }h\sum\limits_{r=0}^{n}{f(a+rh)}$where$h=\dfrac{b-a}{n}$.
We will use the above formula for further simplifications.
Here, we can observe that $\displaystyle \lim_{x \to \infty }\dfrac{1}{n}\left( \sum\limits_{r=0}^{r=n}{{{\left( 1+\dfrac{r}{n} \right)}^{\dfrac{1}{3}}}} \right)$ and $\displaystyle \lim_{n\to \infty }h\sum\limits_{r=0}^{n}{f(a+rh)}$ are similar.
We can compare both of them and write $h=\dfrac{1}{n}$ and$f\left( x \right)={{\left( 1+x \right)}^{\dfrac{1}{3}}}$.
So here it can be written as, $\displaystyle \lim_{n\to \infty }\dfrac{1}{n}\left( \sum\limits_{r=0}^{r=n}{{{\left( 1+\dfrac{r}{n} \right)}^{\dfrac{1}{3}}}} \right)=\int_{0}^{1}{{{\left( 1+x \right)}^{\dfrac{1}{3}}}dx}$
By simplifying this equation we get
$\begin{aligned} & \int_{0}^{1}{{{\left( 1+x \right)}^{\dfrac{1}{3}}}dx=\mathop{\dfrac{{{\left( 1+x \right)}^{\dfrac{4}{3}}}}{\dfrac{4}{3}}}_{x=0}^{x=1}} \\\ & \Rightarrow \dfrac{3}{4}\left( {{2}^{\dfrac{4}{3}}}-1 \right) \\\ \end{aligned}$
Since from the concept of integration we know that, $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C$
By further simplifying this we will get
$\dfrac{3}{4}\left( {{2}^{\dfrac{4}{3}}}-1 \right)=\dfrac{3}{4}{{\left( 2 \right)}^{\dfrac{4}{3}}}-\dfrac{3}{4}$
So we can conclude that,
$\displaystyle \lim_{n\to \infty }\left( \dfrac{{{\left( n+1 \right)}^{\dfrac{1}{3}}}}{{{n}^{\dfrac{4}{3}}}}+\dfrac{{{\left( n+2 \right)}^{\dfrac{1}{3}}}}{{{n}^{\dfrac{4}{3}}}}+...........+\dfrac{{{\left( 2n \right)}^{\dfrac{1}{3}}}}{{{n}^{\dfrac{4}{3}}}} \right)=\dfrac{3}{4}{{\left( 2 \right)}^{\dfrac{4}{3}}}-\dfrac{3}{4}$

So, the correct answer is “Option C”.

Note: Here while we are solving this type of questions we should take care that while performing simplifications and integrations we should not commit any minor or major mistakes. If we even commit a small mistake it will lead us to a different conclusion. We should remember this formula $\int_{a}^{b}{f(x)dx}=\displaystyle \lim_{n\to \infty }h\sum\limits_{r=0}^{n}{f(a+rh)}$ where $h=\dfrac{b-a}{n}$, each and every small bit of this formula is very important. It will be better if we solve the simplifications step by step carefully, probably we will have less chances of getting mistakes in this format.