Question

Find the limiting value of the ratio of the square of the sum of a natural numbers to n times the sum of squares of the n natural number as, n approaches infinity

Answer 1

Sum of the squares of natural numbers = $S_2$

$S_2=1^2+2^2+3^2+4^2+....+n^2$

$S_2=\frac{n(n+1)(2n+1)}{6}$

Sum of natural numbers = $S_1$

$S_1=1+2+3+4+....+n$

$S_1=\frac{n(n+1)}{2}$

Hence,

$\Rightarrow {lim}_{n\to \infty }\frac{{\left[\frac{n(n+1)}{2}\right]}^2}{n\times \left[\frac{n(n+1)(2n+1)}{6}\right]}$

$\Rightarrow {lim}_{n\to \infty }\frac{\frac{n^2(n+1{)}^2}{4}}{\frac{n\times n(n+1)(2n+1)}{6}}$

$\Rightarrow {lim}_{n\to \infty }\frac{3}{2}\frac{(n+1)}{(2n+1)}$

$\Rightarrow {lim}_{n\to \infty }\frac{3}{2}\left(\frac{1+\frac{1}{n}}{2+\frac{1}{n}}\right)$

$\Rightarrow \frac{3}{4}=0.75$