The value of displaystyle limn → ∞ (√n+1+√n+2+....+√2n-1/n3/2) is

Question

The value of displaystyle limn → ∞ (√n+1+√n+2+....+√2n-1/n3/2) is (A) (2/3)(2√2-1) (B) (2/3)(√2-1) (C) (2/3)(√2+1) (D) (2/3)(2√2+1)

Tardigrade Admin · Accepted Answer

displaystyle limn → ∞((√n+1+√n+2+√n+3+......+√2n-1/n3/2)) = displaystyle limn → ∞(√1+(1/n)+√1+(2/√n)+ ..... +√1+(n-1/n)) (1/n) = displaystyle limn → ∞ displaystyle ∑n=1r-1(1/n)√1+(r/n) =∫10√1+xdx=(2/3).(2√2-1)

Q. The value of $n \to \infty lim$ ${\frac{n + 1 + n + 2 + .... + 2 n - 1}{n ^{3/2}}}$ is

$\frac{2}{3} (22 - 1)$

$\frac{2}{3} (2 - 1)$

$\frac{2}{3} (2 + 1)$

$\frac{2}{3} (22 + 1)$

Q. The value of n→∞lim​{n3/2n+1​+n+2​+....+2n−1​​} is

32​(22​−1)

32​(2​−1)

32​(2​+1)

32​(22​+1)

n→∞lim​(n3/2n+1​+n+2​+n+3​+......+2n−1​​) =n→∞lim​(1+n1​​+1+n​2​​+.....+1+nn−1​​)n1​ =n→∞lim​ n=1∑r−1​n1​1+nr​​ =∫01​1+x​dx=32​.(22​−1)

Q. The value of $n \to \infty lim$ ${\frac{n + 1 + n + 2 + .... + 2 n - 1}{n ^{3/2}}}$ is

$\frac{2}{3} (22 - 1)$

$\frac{2}{3} (2 - 1)$

$\frac{2}{3} (2 + 1)$

$\frac{2}{3} (22 + 1)$