displaystyle lim n arrow ∞[(1/1+n)+(1/2+n)+(1/3+n)+ ldots+(1/2 n)] is equal to

Question

displaystyle lim n arrow ∞[(1/1+n)+(1/2+n)+(1/3+n)+ ldots+(1/2 n)] is equal to (A)  log e 2 (B)  log e((2/3)) (C) 0 (D)  log e((3/2))

Tardigrade Admin · Accepted Answer

displaystyle lim n arrow ∞((1/1+n)+ ldots+(1/n+n))= displaystyle lim n arrow ∞ displaystyle∑r=1n (1/n+r) = displaystyle lim n arrow ∞ ∑r=1n (1/n)((1/1+(r)n)) =∫ limits01 (1/1+x) d x=[ℓ ln (1+x]01=ℓ n 2.

Q. $n \to \infty lim [\frac{1}{1 + n} + \frac{1}{2 + n} + \frac{1}{3 + n} + \dots + \frac{1}{2 n}]$ is equal to

$lo g_{e} 2$

$lo g_{e} (\frac{2}{3})$

0

$lo g_{e} (\frac{3}{2})$

$n \to \infty lim (\frac{1}{1 + n} + \dots + \frac{1}{n + n}) = n \to \infty lim r = 1 \sum n \frac{1}{n + r}$
$= n \to \infty lim r = 1 \sum n \frac{1}{n} (\frac{1}{1 + \frac{r}{n}})$
$= 0 \int 1 \frac{1}{1 + x} d x = [ℓ ln (1 + x]_{0}^{1} = ℓ n 2$

Q. n→∞lim​[1+n1​+2+n1​+3+n1​+…+2n1​] is equal to

loge​2

loge​(32​)

0

loge​(23​)

n→∞lim​(1+n1​+…+n+n1​)=n→∞lim​r=1∑n​n+r1​ =n→∞lim​r=1∑n​n1​(1+nr​1​) =0∫1​1+x1​dx=[ℓln(1+x]01​=ℓn2

Q. $n \to \infty lim [\frac{1}{1 + n} + \frac{1}{2 + n} + \frac{1}{3 + n} + \dots + \frac{1}{2 n}]$ is equal to

$lo g_{e} 2$

$lo g_{e} (\frac{2}{3})$

$lo g_{e} (\frac{3}{2})$

$n \to \infty lim (\frac{1}{1 + n} + \dots + \frac{1}{n + n}) = n \to \infty lim r = 1 \sum n \frac{1}{n + r}$
$= n \to \infty lim r = 1 \sum n \frac{1}{n} (\frac{1}{1 + \frac{r}{n}})$
$= 0 \int 1 \frac{1}{1 + x} d x = [ℓ ln (1 + x]_{0}^{1} = ℓ n 2$