Prove the following

Question:

$\lim _{n \rightarrow \infty}\left(\frac{(n+1)^{1 / 3}}{n^{4 / 3}}+\frac{(n+2)^{1 / 3}}{n^{4 / 3}}+\ldots . .+\frac{(2 n)^{1 / 3}}{n^{4 / 3}}\right)$ is equal to :

 

  1. (1) $\frac{3}{4}(2)^{4 / 3}-\frac{3}{4}$

  2. (2) $\frac{4}{3}(2)^{4 / 3}$

  3. (3) $\frac{3}{2}(2)^{4 / 3}-\frac{4}{3}$

  4. (4) $\frac{4}{3}(2)^{3 / 4}$


Correct Option: 1

Solution:

$\lim _{n \rightarrow \infty} \frac{(n+1)^{\frac{1}{3}}+(n+2)^{\frac{1}{3}}+\ldots+(n+n)^{\frac{1}{3}}}{n(n)^{\frac{1}{3}}}$

$=\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{(n+r)^{\frac{1}{3}}}{n \cdot n^{\frac{1}{3}}}$

$=\int_{0}^{1}(1+x)^{\frac{1}{3}} d x$ $\left[\because \frac{\mathrm{r}}{\mathrm{n}} \rightarrow \mathrm{x}\right.$ and $\left.\frac{1}{\mathrm{n}} \rightarrow \frac{\mathrm{d}}{\mathrm{x}}\right]$

$=\left[\frac{3}{4}(1+x)^{\frac{4}{3}}\right]_{0}^{1}=\frac{3}{4}(2)^{\frac{4}{3}}-\frac{3}{4}$

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