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

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

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

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Limits

n \to \infty lim (\frac{( n + 1 ) ^{1/3}}{n ^{4/3}} + \frac{( n + 2 ) ^{1/3}}{n ^{4/3}} + \dots . + \frac{( 2 n ) ^{1/3}}{n ^{4/3}})

is equal to:

Answer: (A)

Rewrite the sum as

lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n} (1 + \frac{r}{n})^{1/3}

, which is the Riemann sum for

\int_{0}^{1} (1 + x)^{1/3} d x

. Evaluating,

\int_{0}^{1} (1 + x)^{1/3} d x = [\frac{3}{4} (1 + x)^{4/3}]_{0}^{1} = \frac{3}{4} (2^{4/3} - 1)

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option

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