Question

${lim}_{n\to \infty }\frac{n^2}{1+2+3+....+n}$

Answer 1

${lim}_{n\to \infty }\frac{n^2}{1+2+3+....+n}$

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

$[\because 1+2+3+...+n=\frac{n(n+1)}{2}]$

$={lim}_{n\to \infty }\frac{2n^2}{n^2+n}=2{lim}_{n\to \infty }\frac{n^2}{n^2+n}$

$=2{lim}_{n\to \infty }\frac{n^2}{(1+\frac{1}{n})}=2\times \frac{1}{1+0}=2$