Question

If $\sum _{r=1}^n{a}_r=\sum _{k=1}^n\sum _{j=1}^k\sum _{i=1}^{j}2,$ then $\underset{m\to \infty }{lim}{\left(\sum _{r=1}^n\frac{1}{a_r}\right)}^m$ is

• A. $0$
• B. $\infty$
• C. $1$
• D. does not exist

Answer 1

Answer: (A)

$0$

Let, $S_n=\sum _{r=1}^n{a}_r=\sum _{k=1}^n\sum _{j=1}^k\sum _{i=1}^{j}2$
$\Rightarrow S_n=2\sum _{k=1}^n\sum _{j=1}^{k}j=\sum _{k=1}^{n}k(k+1)$
$\Rightarrow S_n=\sum _{k=1}^n{k}^2+\sum _{k=1}^{n}k=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$
$\Rightarrow S_n=\frac{n(n+1)(n+2)}{3}$
We know that, $a_r=S_n-S_{n-1}$
Therefore, $a_r=\frac{1}{3}\left\{r\right(r+1\left)\right(r+2)-(r-1\left)r\right(r+1\left)\right\}=r(r+1)$
$\Rightarrow \frac{1}{a_r}=\frac{1}{r(r+1)}=\frac{1}{r}-\frac{1}{r+1}$
$\Rightarrow \sum _{r=1}^n\frac{1}{a_r}=1-\frac{1}{n+1}=\frac{n}{n+1}$
Since, $\frac{n}{n+1}<1$
Therefore, $\underset{m\to \infty }{lim}{\left(\sum _{r=1}^n\frac{1}{a_r}\right)}^m=\underset{m\to \infty }{lim}{\left(\frac{n}{n+1}\right)}^m=0$
Ans: A