displaystyle limn → ∞[(1/2n+1)+(1/2n+2)+.....+(1/2n+n)]=

Question

displaystyle limn → ∞[(1/2n+1)+(1/2n+2)+.....+(1/2n+n)]= (A) loge=(1/3) (B) loge=(2/3) (C) loge=(3/2) (D) loge=(4/3)

Tardigrade Admin · Accepted Answer

displaystyle limn → ∞[(1/2n+1)+(1/2n+2)+.....+(1/2n+n)] = displaystyle limn → ∞[(1/2+(1)n)+(1/2+(2)n)+...+(1/2+(n)n)] = displaystyle limn → ∞(1/n) displaystyle ∑n=1n(1/2+(2)n)=∫ limits10(dx/2+x) =[log(2+x)]10]=log 3-log 2=log (3/2)

Q. $n \to \infty lim$ $[\frac{1}{2 n + 1} + \frac{1}{2 n + 2} + ..... + \frac{1}{2 n + n}] =$

$l o g_{e} = \frac{1}{3}$

$l o g_{e} = \frac{2}{3}$

$l o g_{e} = \frac{3}{2}$

$l o g_{e} = \frac{4}{3}$

Q. n→∞lim​[2n+11​+2n+21​+.....+2n+n1​]=

loge​=31​

loge​=32​

loge​=23​

loge​=34​

n→∞lim​[2n+11​+2n+21​+.....+2n+n1​] =n→∞lim​[2+n1​1​+2+n2​1​+...+2+nn​1​] =n→∞lim​n1​n=1∑n​2+n2​1​=0∫1​2+xdx​ =[log(2+x)]01​]=log3−log2=log23​

Q. $n \to \infty lim$ $[\frac{1}{2 n + 1} + \frac{1}{2 n + 2} + ..... + \frac{1}{2 n + n}] =$

$l o g_{e} = \frac{1}{3}$

$l o g_{e} = \frac{2}{3}$

$l o g_{e} = \frac{3}{2}$

$l o g_{e} = \frac{4}{3}$