displaystyle lim n arrow ∞ (1/n+m)+(1/n+2 m)+(1/n+3 m)+ ldots+(1/n+n m) =

Question

displaystyle lim n arrow ∞ (1/n+m)+(1/n+2 m)+(1/n+3 m)+ ldots+(1/n+n m) = (A) ( log 0(m)/m) (B) ( log 0(1+m)/1+m) (C) ( log 0(1+m)/m) (D) ( log 0(1+m)/1-m)

Tardigrade Admin · Accepted Answer

displaystyle lim n arrow ∞ (1/n+m)+(1/n+2 m)+(1/n+3 m)+ ldots+(1/n+n m) = displaystyle lim n arrow ∞ (1/n) beginarrayc(1/1+(m)n)+(1/1+2((m)n))+(1/1+3((m)n))+ ldots . +(1/1+n((m)n)) endarray = displaystyle lim n arrow ∞ (1/n)(∑k=1n (1/1+m((k)n))) =∫ limits01 (1/1+m x) d x=(1/m) ∫ limits01 (m/1+m x) d x =(1/m)[ log e(1+m x)]01=(1/m)[ log e(1+m)- log (1)] =( log e(1+m)/m)

Q. $n \to \infty lim {\frac{1}{n + m} + \frac{1}{n + 2 m} + \frac{1}{n + 3 m} + \dots + \frac{1}{n + nm}} =$

$\frac{l o g _{0} ( m )}{m}$

$\frac{l o g _{0} ( 1 + m )}{1 + m}$

$\frac{l o g _{0} ( 1 + m )}{m}$

$\frac{l o g _{0} ( 1 + m )}{1 - m}$

Q. n→∞lim​{n+m1​+n+2m1​+n+3m1​+…+n+nm1​}=

mlog0​(m)​

1+mlog0​(1+m)​

mlog0​(1+m)​

1−mlog0​(1+m)​

Q. $n \to \infty lim {\frac{1}{n + m} + \frac{1}{n + 2 m} + \frac{1}{n + 3 m} + \dots + \frac{1}{n + nm}} =$

$\frac{l o g _{0} ( m )}{m}$

$\frac{l o g _{0} ( 1 + m )}{1 + m}$

$\frac{l o g _{0} ( 1 + m )}{m}$

$\frac{l o g _{0} ( 1 + m )}{1 - m}$