Question

Let $\left[y\right]$ denote the greatest integer less than or equal to $y.$ If $f:(0,\infty )\to N$ is defined by $f\left(x\right)=\left[\frac{x^2+x+1}{x^2+1}\right]+\left[\frac{4x^2+x+2}{2x^2+1}\right]+\left[\frac{9x^2+x+3}{3x^2+1}\right]+\cdots +\left[\frac{n^2{x}^2+x+n}{n{x}^2+1}\right]$ for $n\in N,$ then the value of $\underset{n\to \infty }{lim}\left(\frac{f\left(x\right)-n}{{\left(f\right(x\left)\right)}^2-\frac{n^3(n+2)}{4}}\right)$ is

Answer 1

$f\left(x\right)=\left[\frac{x^2+x+1}{x^2+1}\right]+\left[\frac{4x^2+x+2}{2x^2+1}\right]+\left[\frac{9x^2+x+3}{3x^2+1}\right]+\cdots +\left[\frac{n^2{x}^2+x+n}{n{x}^2+1}\right]=[1+\frac{x}{x^2+1}]+[2+\frac{x}{2x^2+1}]+[3+\frac{x}{3x^2+1}]+\cdots +[n+\frac{x}{n{x}^2+1}]=1+2+\cdots +n+\left[\frac{x}{x^2+1}\right]+\left[\frac{x}{2x^2+1}\right]+\left[\frac{x}{3x^2+1}\right]+\cdots +\left[\frac{x}{n{x}^2+1}\right]$

For $x>0,$
$\frac{x}{n{x}^2+1}\in (0,1)$
$\therefore \left[\frac{x}{x^2+1}\right]=\left[\frac{x}{2x^2+1}\right]=\cdots =\left[\frac{x}{n{x}^2+1}\right]=0$
$f\left(x\right)=\frac{n(n+1)}{2}$

Now,
$\underset{n\to \infty }{lim}\left(\frac{f\left(x\right)-n}{\left(f\right(x){)}^2-\frac{n^3(n+2)}{4}}\right)=\underset{n\to \infty }{lim}\left(\frac{\frac{n(n+1)}{2}-n}{\frac{n^2(n+1{)}^2}{4}-\frac{n^3(n+2)}{4}}\right)=2\underset{n\to \infty }{lim}\left(\frac{n(n-1)}{n^2}\right)=2\underset{n\to \infty }{lim}(1-\frac{1}{n})=2$