Question

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

Answer 1

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

$\Rightarrow (\frac{n(n+1)}{2}+2n).(\frac{n(n-1)}{2}+3(n-1\left)\right).(\frac{(n-2)(n-1)}{2}+4(n-2\left)\right).(\frac{(n-3)(n-2)}{2}+5(n-3\left)\right).\left(\frac{(n-(n-1\left)\right)\times (n-(n-2\left)\right)}{2}\right)+(2+n-1)\times (n-(n-1\left)\right).\left(\frac{(\frac{2}{2}+(n+1\left)\right)}{n^4}\right)$

$\Rightarrow \left(\frac{(\frac{(n+1)}{2}+2).\left(\frac{n^2+3n-6}{2}\right)......(n+2)}{n^3}\right)$ taking $n$ common

$\Rightarrow \frac{1}{2}\times 1.\left(\frac{(n-2)(n-17)}{2}\right).(n-3)(n-32).....(n+2)$

$=\infty$