Question:- Find Limit $$L=\lim_{n\to \infty }\frac{1}{n}\sum_{k=1}^{n}\left\lfloor 2\sqrt{\frac{n}{k}} \right\rfloor -2\left\lfloor\sqrt{\frac{n}{k}} \right\rfloor \text , $$ where $\lfloor x \rfloor$ represents greatest integer function.
Yesterday, my friend sent me this limit question.Greatest integer function is the biggest problem here.I don't know how to evaluate the summation to find the given limit.
Can anybody help me!!
Best Answer
As this is a Riemann sum, you can convert it into an integral.
This becomes:
$$\int_0^1 \left \lfloor \frac2{\sqrt x} \right \rfloor -2\left \lfloor\frac1{\sqrt x} \right \rfloor\,dx$$ Put $\sqrt x \rightarrow 1/t$ to get: $$ = 2\int_1^\infty \frac{\left \lfloor 2t \right \rfloor}{t^3} -2\frac{\left \lfloor t \right \rfloor}{t^3}\,dt$$ $$ = 2\left(\sum_{r=1}^\infty\int_{(r+1)/2}^{r/2 + 1}\frac{r+1}{t^3}\,dt - 2\sum_{r=1}^\infty\int_{r}^{r + 1}\frac{r}{t^3}\,dt\right)$$ $$ = 2\sum_{r=1}^\infty\left(\frac{2(2r+3)}{(1+r)(2+r)^2} - \frac{2r+1}{r(1+r)^2}\right)$$
$$ = 2\sum_{r=1}^\infty\left(\frac{4}{(r+1)(r+2)} - \frac{2}{(r+1)(r+2)^2} - \frac{2}{r(r+1)} + \frac{1}{r(1+r)^2}\right)$$ $$ = 2\sum_{r=1}^\infty\left(\frac{1}{r(1+r)^2}-\frac{2}{(r+1)(r+2)^2}\right)$$ $$ = 1 - 2\sum_{r=1}^\infty\left(\frac{1}{r(1+r)^2}\right)$$ $$= \boxed{\frac{\pi^2}3 - 3}$$