Question

$\underset{n\to \infty }{lim}\frac{(1^2+2^2+...+n^2)\sqrt[n]{nn}}{(n+1)(n+10)(n+100)}=$

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

Answer 1

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

We know that $1^2+2^2+3^2+4^2+......+n^2=\frac{n(n+1)(2n+1)}{6}$

$\underset{n\to \infty }{lim}\sqrt[n]{nn}=\underset{n\to \infty }{lim}{e}^{{\log }_e\sqrt[n]{nn}}=\underset{n\to \infty }{lim}{e}^{\frac{{\log }_{e}n}{n}}$

As we know that graph of ${\log }_{e}n$ is always below $n$

$\therefore \underset{n\to \infty }{lim}\frac{{\log }_{e}n}{n}=0$

$⟹\underset{n\to \infty }{lim}\sqrt[n]{nn}=1$
then,

$\therefore \underset{n\to \infty }{lim}\frac{(1^2+2^2+3^2....+n^2)\left(\sqrt[n]{nn}\right)}{(n+1)(n+10)(n+100)}=\underset{n\to \infty }{lim}\frac{n(n+1)(2n+1)}{6(n+1)(n+10)(n+100)}$

$=\underset{n\to \infty }{lim}\frac{1(1+\frac{1}{n})(2+\frac{1}{n})}{6(1+\frac{1}{n})(1+\frac{10}{n})(1+\frac{100}{n})}=\frac{2}{6}=\frac{1}{3}$

$\therefore$ The value of limit equals $\frac{1}{3}$

Hence, option B is correct.