A calculus problem by Sabhrant Sachan

The $mth$ term of the sum is of the form $\dfrac{\displaystyle\sum_{r=1}^{m}r}{\displaystyle\sum_{r=1}^m r^3}=\dfrac{2}{m(m+1)}=2\left(\dfrac 1m-\dfrac{1}{m+1}\right)$ $\small{\color{#3D99F6}{Use~\displaystyle\sum r=\dfrac{m(m+1)}{2},\ \displaystyle\sum r^3=\left(\dfrac{m(m+1)}{2}\right)^2}}$

Thus required sum is:- $\large\displaystyle\sum_{m=1}^{n}2\left(\dfrac 1m-\dfrac{1}{m+1}\right)$ Which is a telescopic series and comes out to be:- $\large S_n=2-\dfrac{2}{n+1}$ $\large\lfloor S_n\rfloor=1~~(\because ~0<\dfrac{2}{n+1}<1)$ $(\text{Recall } \lfloor x\rfloor=1 \text { if } 1<x<2$ Hence, $\large\displaystyle\lim_{n\to\infty}(1)=\boxed{\large 1}$