Question

The value of $\underset{n\to \infty }{lim}\frac{1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n(n+1)}{n^3}$ is

• A. $1$
• B. $-1$
• C. $\frac{1}{3}$
• D. $-\frac{1}{3}$

Answer 1

The correct option is C $\frac{1}{3}$

$\underset{n\to \infty }{lim}\frac{1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n(n+1)}{n^3}$
$=\underset{n\to \infty }{lim}\frac{\sum _{r=1}^{n}r(r+1)}{n^3}$
$=\underset{n\to \infty }{lim}\frac{\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}}{n^3}$
$=\underset{n\to \infty }{lim}\frac{n(n+1)(n+\frac{1}{2})}{3n^3}+\frac{n(n+1)}{2n^3}=\frac{1}{3}+0=\frac{1}{3}$