If $A = \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{999}}+\frac{1}{\sqrt{1000}}.$ Then $\lfloor A \rfloor$ is, 
If $\displaystyle A = \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots\cdots\cdots+\frac{1}{\sqrt{999}}+\frac{1}{\sqrt{1000}}.$
Then $\lfloor A \rfloor$ is, where $\lfloor A\rfloor = A-\{A\}.$

$\bf{My\; Try::}$ Using $$\left(\sqrt{k}+\sqrt{k-1}\right)<2\sqrt{k}<\left(\sqrt{k+1}+\sqrt{k}\right)\;,$$ where $k\in N$ and $k\geq 2$
Then $$\displaystyle \left(\sqrt{k+1}-\sqrt{k}\right)<\frac{1}{2\sqrt{k}}<\left(\sqrt{k}-\sqrt{k-1}\right)$$
So $$\displaystyle 2\sum_{k=2}^{1000}\left(\sqrt{k+1}-\sqrt{k}\right)<\sum_{k=2}^{1000}\frac{1}{\sqrt{k}}<2\sum_{k=2}^{1000}\left(\sqrt{k}-\sqrt{k-1}\right)$$
So we get $$\displaystyle 2(10\sqrt{10}-\sqrt{2}) < A < 2(10\sqrt{10}-1)$$
So we get $$60.41\approx <A<\approx 61.24$$. So $$\lfloor A \rfloor  = 61$$
My Question is how can we solve using Concept of definite Integral, plz explain me
Thanks
 A: Notice that:
$$60.41<60.5<61.24$$
but,
$$\lfloor60.5\rfloor=60\ne61$$
so your logic is wrong.  To correct it, simply note that:
$$\sum_{k=2}^{1000}\frac1{\sqrt k}=\sum_{a=2}^{10}\frac1{\sqrt a}+\sum_{b=11}^{1000}\frac1{\sqrt b}$$
Likewise, you will find that
$$\sum_{a=2}^{10}\frac1{\sqrt a}\approx4.02100$$
$$56.64392\approx2(\sqrt{1001}-\sqrt{11})<\sum_{b=11}^{1000}\frac1{\sqrt b}<2(\sqrt{1000}-\sqrt{10})\approx56.92100$$
Thus,
$$60.66492<\sum_{k=2}^{1000}\frac1{\sqrt k}<60.94200$$
And so we see the correct answer is actually $60$.

Another approach would to brute force it with the Euler-Maclaurin formula:
$$\sum_{k=1}^n\frac1{\sqrt k}=\zeta(1/2)+2\sqrt n+\frac1{2\sqrt n}+\mathcal O(n^{-3/2})$$
where $\zeta(1/2)\approx-1.4603545088$, and this gives us
$$\sum_{k=2}^{1000}\frac1{\sqrt k}\approx-1+\zeta(1/2)+2\sqrt{1000}+\frac1{2\sqrt{1000}}=60.80101$$
with an error at most of $\frac1{24\times1000^{3/2}}$, so we're pretty safe.
A: With $k$ integer and
$$x< k< x+1$$
we have
$$\frac1{\sqrt{x+1}}<\frac1{\sqrt k}<\frac1{\sqrt x}.$$
Then integrating in unit intervals,
$$\int_1^{1000}\frac{dx}{\sqrt{x+1}}<\sum_{k=2}^{1000}\frac1{\sqrt k}<\int_1^{1000}\frac{dx}{\sqrt x},$$ i.e.
$$60.449\approx2(\sqrt{1001}-\sqrt2)<\sum_{k=2}^{1000}\frac1{\sqrt k}<2(\sqrt{1000}-\sqrt1)\approx61.245$$
Unfortunately, this bracketing does not allow us to conclude, as the floor could be $60$ or $61$. We can tighten by pulling the first terms out,
$$\color{green}{60}.520\approx2(\sqrt{1001}-\sqrt3)+\frac1{\sqrt2}<\sum_{k=2}^{1000}\frac1{\sqrt k}<2(\sqrt{1000}-\sqrt2)+\frac1{\sqrt2}\approx\color{green}{61}.124$$
$$\color{green}{60}.561\approx2(\sqrt{1001}-\sqrt4)+\frac1{\sqrt2}+\frac1{\sqrt3}<\sum_{k=2}^{1000}\frac1{\sqrt k}<2(\sqrt{1000}-\sqrt3)+\frac1{\sqrt2}+\frac1{\sqrt3}\approx\color{green}{61}.066$$
$$\color{green}{60}.589\approx2(\sqrt{1001}-\sqrt5)+\frac1{\sqrt2}+\frac1{\sqrt3}+\frac1{\sqrt4}<\sum_{k=2}^{1000}\frac1{\sqrt k}<2(\sqrt{1000}-\sqrt4)+\frac1{\sqrt2}+\frac1{\sqrt3}+\frac1{\sqrt4}\approx\color{green}{61}.030$$
$$\cdots$$
Further terms confirm $60$.
