Question

The value of $\underset{n\to \infty }{lim}\frac{1}{n^3}\{1+3+6+\dots \dots \dots .+\frac{n(n+1)}{2}\}$ is

• A. $0$
• B. $2$
• C. $\frac{1}{6}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $\frac{1}{6}$

The $n^{th}$ term of the sequence is given by, $T_n=\frac{n(n+1)}{2}$

Thus, the sum of the series will be given by,

$S_n=\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4}$

The given limit is,

$\underset{n\to \infty }{lim}\frac{S_n}{n^3}$

$\underset{n\to \infty }{lim}\frac{n(n+1)(n+2)}{6n^3}$

Dividing the numerator and denominator by $n^3$
$\underset{n\to \infty }{lim}\frac{(1+\frac{1}{n})(1+\frac{2}{n})}{6}$

Therefore, applying the limit,
$L=\frac{1}{6}$

Hence, option ${}^{\prime }{C}^{\prime }$is correct.