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

Question

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

MARKS App · Accepted Answer

The given limit can be written as l i m n → ∞ 1 n 1 + 1 n 1 3 + 1 + 2 n 1 3 + … . + 1 + n n 1 3 = l i m n → ∞ 1 n ∑ r = 1 n 1 + r n 1 3 = ∫ 0 1 ( 1 + x ) 1 / 3 d x = 1 + x 4 3 4 3 0 1 = 3 4 ( 2 4 / 3 – 1 ) = 3 4 . 2 4 / 3 - 3 4

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