Question

${lim}_{x\to \infty }\left[\frac{1n+2(n-1)+\dots +n.1}{1^3+2^3+\dots +n^3}\right]$ is equal to

• A. $\frac{2}{3}$
• B. $0$
• C. $e^{\frac{1}{2}}$
• D. $e^{\frac{2}{3}}$

Answer 1

The correct option is B $0$

$\underset{x\to \infty }{lim}$ $\left[\frac{1n+2(n-1)+...+n.1}{1^3+2^3+3^3+...n^3}\right]$

the numberator can be simplified as

$1.n+2.(n-1)...+n.1$

$={\sum }_{k=1}^{n}k.(n-(k-1))$

$={\sum }_{k=1}^{n}k(n-k-1)={\sum }_{k=1}^{n}nk-{\sum }_{k=1}^n{k}^2-{\sum }_{k=1}^{n}k$

$=n{\sum }_{k=1}^{n}k-{\sum }_{k=1}^n{k}^2-{\sum }_{k=1}^{n}k=\frac{n\left(n(n+1)\right)}{2}-\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}$

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

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

also $1^3+2^3+3^3...n^3=\frac{n^2{(n+1)}^2}{4}$

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

$=\underset{n\to \infty }{lim}\frac{\frac{n-4}{3}}{\frac{n(n+1)}{2}}$

$=\underset{n\to \infty }{lim}\frac{2}{3}\left[\frac{n-4}{n(n+1)}\right]$

$=\underset{n\to \infty }{lim}\frac{2}{3}\left[\frac{1-\frac{4}{n}}{1+n}\right]$

$=0$

None of the options are correct.