Question

The value of $\underset{x\to \infty }{lim}\frac{1\cdot \sum _{r=1}^{n}r+2\cdot \sum _{r=1}^{n-1}r+3\cdot \sum _{r=1}^{n-2}r+....+n\cdot 1}{n^4}$

• A. $\frac{1}{24}$
• B. $\frac{1}{12}$
• C. $\frac{1}{6}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{24}$

Given: ${lim}_{n\to \infty }\frac{{\sum }_{r=1}^{n}r+2{\sum }_{r=1}^{n-1}r+...n.1}{n^4}$

To find: the value of the limit

Sol: We can write the above expression as

${lim}_{n\to \infty }\frac{{\sum }_{k=1}^n\left(k{\sum }_{r=1}^{n-k+1}r\right)}{n^4}\therefore {\sum }_{k=1}^n\left(k{\sum }_{r=1}^{n+1-k}r\right)={\sum }_{k=1}^{n}k\frac{(n+1-k)(n+2-k)}{2}={\sum }_{k=1}^{n}k\left\{\frac{n^2+k^2-2nk+3n-3k+2}{2}\right\}={\sum }_{k=1}^{n}k\left\{\frac{n^2+k^2-(2n+3)k+(3n+2)}{2}\right\}$

$=\left(\frac{2+n^2+3n}{2}\right){\sum }_{k=1}^{n}k+\frac{1}{2}{\sum }_{k=1}^n{k}^3-\frac{(2n+3)}{2}{\sum }_{k=1}^n{k}^2=\left(\frac{2+n^2+3n}{2}\right)\times \frac{n(n+1)}{2}+\frac{1}{2}{\left(\frac{n(n+1)}{2}\right)}^2-\frac{(2n+3)}{2}\frac{n(n+1)(2n+1)}{6}=\frac{n(n+1)}{2}[\frac{2+n^2+3n}{2}+\frac{n^2+n}{4}-\frac{{4n}^2+8n+3}{6}]=\frac{n(n+1)(n^2+5n+6)}{24}$

$\therefore {lim}_{n\to 0}\frac{{\sum }_{k=1}^n\left(k{\sum }_{r=1}^{n-k+1}r\right)}{n^4}=\frac{1}{24}{lim}_{n\to \infty }\frac{n(n+1)(n+2)(n+3)}{n^4}=\frac{1}{24}{lim}_{n\to \infty }(1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})=\frac{1}{24}$