Question

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

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{6}$
• D. $\frac{1}{8}$

Answer 1

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

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

$=\underset{n\to \infty }{lim}\frac{\sum _{m=1}^n\frac{m(m+1)}{2}}{n^3}$

$=\underset{n\to \infty }{lim}\frac{\sum _{m=1}^n{m}^2+m}{2n^3}$

$=\frac{1}{2}\underset{n\to \infty }{lim}\frac{\sum _{m=1}^n{m}^2+\sum _{m=1}^{n}m}{n^3}$

It is known that:

$\sum _{m=1}^{n}m=\frac{n(n+1)}{2}$

$\sum _{m=1}^n{m}^2=\frac{n(n+1)(2n+1)}{6}$

Hence,

$\frac{1}{2}\underset{n\to \infty }{lim}\frac{\sum _{m=1}^n{m}^2+\sum _{m=1}^{n}m}{n^3}$

$=\underset{n\to \infty }{lim}\frac{1}{\left(2n^3\right)}[\frac{n(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}]$

$=\underset{n\to \infty }{lim}\frac{1}{\left(2n^3\right)}[\frac{n(n+1)}{2}]+\underset{n\to \infty }{lim}\frac{1}{\left(2n^3\right)}[\frac{n(n+1)(2n+1)}{6}]$

In case of $\underset{n\to \infty }{lim}$

if the degree of the polynomial in the numerator $<$ degree of the denominator, the limit is $0$.

If the degrees of the numerator and denominator are equal, the limit is the ratio of the leading terms.

Hence,

$=\underset{n\to \infty }{lim}\frac{1}{\left(2n^3\right)}[\frac{n(n+1)}{2}]+\underset{n\to \infty }{lim}\frac{1}{\left(2n^3\right)}[\frac{n(n+1)(2n+1)}{6}]$

$=0+\frac{2}{2\times 6}$

$=\frac{1}{6}$