Question

Evaluate:

${lim}_{n\to \infty }\frac{1.2+2.3+3.4+...+n(n+1)}{n^3}$

Answer 1

${lim}_{n\to \infty }\frac{1.2+2.3+3.4+...+n(n+1)}{n^3}$

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

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

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

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

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

$=\frac{1}{3}{lim}_{n\to \infty }(1+\frac{1}{n})(1+\frac{2}{n})asx\to \infty ,\frac{1}{x}\to 0$

$=\frac{1}{3}(1+0)\times (1+0)$

$=\frac{1}{3}$