$\lim _{n ightarrow \infty}\left[\begin{array}{l}\frac{1}{n}+\frac{n^2}{(n+1)^3}+\frac{n^2}{(n+2)^3}+\frac{n^2}{(n+3)^3} \ +\ldots+\frac{1}{125 n}\end{array} ight]=$

Question

$\lim _{n ightarrow \infty}\left[\begin{array}{l}\frac{1}{n}+\frac{n^2}{(n+1)^3}+\frac{n^2}{(n+2)^3}+\frac{n^2}{(n+3)^3} \ +\ldots+\frac{1}{125 n}\end{array}ight]=$

Aicharya · Accepted Answer

Given, $ \begin{aligned} & \lim _{n ightarrow \infty}\left[\frac{1}{n}+\frac{n^2}{(n+1)^3}+\frac{n^2}{(n+2)^3}+\frac{n^2}{(n+3)^3}+\ldots \frac{1}{125 n}ight] \ & =\lim _{n ightarrow \infty}\left[\begin{array}{l} \frac{n^2}{(n+0)^3}+\frac{n^2}{(n+1)^3}+\frac{n^2}{(n+2)^3}+\frac{n^2}{(n+3)^3} \ +\ldots+\frac{n^2}{(n+4 n)^3} \end{array}ight] \ & =\lim _{n ightarrow \infty} \sum_{r=0}^{4 n} \frac{n^2}{(n+r)^3} \end{aligned} $ $ \begin{aligned} & 	ext { Let } \frac{r}{n}=x 	ext { and } \frac{1}{n}=d x=\int_0^4 \frac{d x}{(1+x)^3}=\left[\frac{(1+x)^{-2}}{-2}ight]_0^4 \ & =-\frac{1}{2}\left[\frac{1}{5^2}-1ight]=\frac{1}{2}\left(1-\frac{1}{25}ight)=\frac{12}{25} . \end{aligned} $ Hence, option (c) is correct

$\lim _{n \rightarrow…

Solution