mathop {lim }limits_{n to infty } left( {frac | vedclass

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(A) The given expression can be written as: $\mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^n \frac{(n+r)^{1/3}}{n^{4/3}} = \mathop {\lim

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given expression can be written as: $\mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^n \frac{(n+r)^{1/3}}{n^{4/3}} = \mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^n \frac{1}{n} \left( \frac{n+r}{n} ight)^{1/3}$ Using the definition of a definite integral as the limit of a sum: $\int\limits_0^1 (1+x)^{1/3} dx$ Evaluating the integral: $= \left[ \frac{(1+x)^{4/3}}{4/3} ight]_0^1 = \frac{3}{4} \left( (1+1)^{4/3} - (1+0)^{4/3} ight)$ $= \frac{3}{4} (2^{4/3} - 1) = \frac{3}{4} 2^{4/3} - \frac{3}{4}$

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

Similar Questions

For $a \in \mathbb{R}$ (the set of all real numbers),$a \neq -1$,if $\lim_{n \to \infty} \frac{1^a + 2^a + \dots + n^a}{(n+1)^{a-1}[(na+1) + (na+2) + \dots + (na+n)]} = \frac{1}{60}$,then $a$ is equal to:

Let $S_n = \sum_{k=1}^n \frac{n}{n^2+kn+k^2}$ and $T_n = \sum_{k=0}^{n-1} \frac{n}{n^2+kn+k^2}$ for $n=1, 2, 3, \ldots$. Then,

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

$\lim _{n \rightarrow \infty}\left[\frac{\sqrt{n^2-1^2}}{n^2}+\frac{\sqrt{n^2-2^2}}{n^2}+\frac{\sqrt{n^2-3^2}}{n^2}+\ldots+\frac{\sqrt{n^2-n^2}}{n^2}\right]=$

$\lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n}\right]=$

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