Question

${\sum }_{t=1}^n{t}_r=\frac{1}{6}n(n+1)(n+2)\forall n\ge 1.$ Then ${lim}_{n\to \infty }{\sum }_{r=1}^n\frac{1}{t_r}$ is

• A. $2$
• B. $3$
• C. $\frac{3}{2}$
• D. $6$

Answer 1

Answer: (A)

$2$

We have,

${\sum }_{r=1}^n{t}_r=\frac{1}{6}n\left(n+1\right)\left(n+2\right)t_{n+1}=\frac{1}{6}\left(n+1\right)\left(n+2\right)\left(n+3\right)-\frac{1}{6}n\left(n+1\right)\left(n+2\right)t_{n+1}=\frac{1}{6}\left(n+1\right)\left(n+2\right)\left[\left(n+3\right)-n\right]=\frac{1}{2}\left(n+1\right)\left(n+2\right)t_n=\frac{1}{2}n\left(n+1\right){\sum }_{r=1}^n\frac{1}{t_r}={\sum }_{r=1}^n\frac{2}{r\left(r+1\right)}={lim}_{n\to \infty }{\sum }_{r=1}^{n}2\left[\frac{1}{r}-\frac{1}{r+1}\right]={lim}_{n\to \infty }\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+........+\left(\frac{1}{n}-\frac{1}{n+1}\right)\right]={lim}_{n\to \infty }2\left[1-\frac{1}{n+1}\right]=2$

$Hence,optionAiscorrectanswer.$