Question

${lim}_{n\to \infty }\frac{1^3+2^3+3^3+....+n^3}{(n-1{)}^4}$

Answer 1

${lim}_{n\to \infty }\frac{1^3+2^3+3^3+....+n^3}{(n-1{)}^4}$

$={lim}_{n\to \infty }\frac{{\left[\frac{n(n+1)}{2}\right]}^2}{{\left[n(1-\frac{1}{n})\right]}^4}$

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

$={lim}_{n\to \infty }\frac{\frac{n^2(n+1{)}^2}{4}}{n^4{(1-\frac{1}{n})}^4}=\frac{1}{4}$

$={lim}_{n\to \infty }\frac{n^4{(1+\frac{1}{n})}^2}{n^4{(1-\frac{1}{n})}^4}$

$=\frac{1}{4}\frac{{(1+\frac{1}{n})}^2}{{(1-\frac{1}{n})}^4}$

$asn\to \infty ,\frac{1}{n}\to 0$

$=\frac{1}{4}\left(\frac{1+0}{1-0}\right)$

$=\frac{1}{4}$