# Find $L=\lim_{n\to \infty }\frac{1}{n}\sum_{k=1}^{n}\left\lfloor 2\sqrt{\frac{n}{k}} \right\rfloor -2\left\lfloor \sqrt{\frac{n}{k}} \right\rfloor$

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!!

• Convert this into a Riemann sum, then: math.stackexchange.com/questions/3749611/… Commented Jul 8, 2020 at 17:39
• @Nikunj I don't see how to convert it to a Riemann sum. The interval of integration would go from $1$ to $n$. Commented Jul 8, 2020 at 17:56
• No, $x = k/n$, would go from $0$ to $1$ as $k$ goes from $1$ to $n$ Commented Jul 8, 2020 at 18:00
• @Nikunj Oh, duh. Yes, you're right. Commented Jul 8, 2020 at 18:09

$$\int_0^1 \left \lfloor \frac2{\sqrt x} \right \rfloor -2\left \lfloor\frac1{\sqrt x} \right \rfloor\,dx$$ Put $$\sqrt x \rightarrow 1/t$$ to get: $$= 2\int_1^\infty \frac{\left \lfloor 2t \right \rfloor}{t^3} -2\frac{\left \lfloor t \right \rfloor}{t^3}\,dt$$ $$= 2\left(\sum_{r=1}^\infty\int_{(r+1)/2}^{r/2 + 1}\frac{r+1}{t^3}\,dt - 2\sum_{r=1}^\infty\int_{r}^{r + 1}\frac{r}{t^3}\,dt\right)$$ $$= 2\sum_{r=1}^\infty\left(\frac{2(2r+3)}{(1+r)(2+r)^2} - \frac{2r+1}{r(1+r)^2}\right)$$
$$= 2\sum_{r=1}^\infty\left(\frac{4}{(r+1)(r+2)} - \frac{2}{(r+1)(r+2)^2} - \frac{2}{r(r+1)} + \frac{1}{r(1+r)^2}\right)$$ $$= 2\sum_{r=1}^\infty\left(\frac{1}{r(1+r)^2}-\frac{2}{(r+1)(r+2)^2}\right)$$ $$= 1 - 2\sum_{r=1}^\infty\left(\frac{1}{r(1+r)^2}\right)$$ $$= \boxed{\frac{\pi^2}3 - 3}$$