$\left\lfloor \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}\right\rfloor =?$ $$\left\lfloor \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}\right\rfloor =?$$ 
I can't find $$? \leq  \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}} \leq ?$$ I remeber there was a sharp bound , Am i right  ?
I am thankful if someone help me to find the bound  , or give another idea
Last question : Is it possible to find $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}$  by integral  ?
 A: Use
$$\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\in\left(\frac{1}{2\sqrt{n+1}}, \frac{1}{2\sqrt{n}}\right)$$
or, to put it differently:
$$2(\sqrt{n+1}-\sqrt{n})\lt\frac{1}{\sqrt{n}}\lt2(\sqrt{n}-\sqrt{n-1})$$
so you can squeeze your sum between two telescopic sums.
A: By the EML formula
$$ H_n^{(1/2)} = 2\sqrt{n}+\zeta\left(\tfrac{1}{2}\right)+\frac{1}{2\sqrt{n}}+O\left(\frac{1}{n^{3/2}}\right) $$
hence $H_{1024}^{(1/2)}\approx \color{red}{62}+\frac{1}{2}$. There are many questions on MSE asking for approximations of $H_n^{(1/2)}$: the Hermite-Hadamard inequality/the trapezoid method for the estimation of $\int_{1}^{1024}\frac{dx}{\sqrt{x}}$ provide straightforward answers for the task at hand.
A: Since $$\frac1{\sqrt{n+1}}<2(\sqrt{n+1}-\sqrt{n})=\frac2{\sqrt{n+1}+\sqrt{n}}<\frac1{\sqrt{n}},$$
we have $$2(\sqrt{n+1}-\sqrt{2})<\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{1}),$$ i.e.
$$2\sqrt{n+1}-2\sqrt{2}+1<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}<2\sqrt{n}-1.$$
For $n=1024$, the bounds are $2\sqrt{1025}-2\sqrt{2}+1\approx62.2$ and $63$, so the result would be $62$. 
A: To answer your last question: no.
Say we got $f(x) = 2x$, the primitive function would be $F(x) = x^2$. Now say we want to take the integral for $[0, 10]$ then we'd get 
$$\int_{0}^{10} f(x) dx = F(10) - F(0) = 10^2 = 100$$ 
However, if we'd only take integers (like in your example) the answer would be 
$$\sum_{x = 0}^{10}f(x) = 0 + 2 + 4 + 6 + 8 + ... + 20 = 110$$
So like, where in your question you actually got the formula $$\sum_{n=1}^{1024}\frac{1}{\sqrt{n}}$$ You can't treat it like it's the integral of $g(n) = \frac{1}{\sqrt{n}}$ because that'd give the result $$\int_{1}^{1024} g(n)dn = G(1024) - G(1) = 2 * \sqrt{1024} - 2 * \sqrt{1} = 2 * 64 - 2 = 62$$
Which does not, even though it seems like it gets very, very close, equal the summation you started with (Thanks a lot to @AlexS for correcting my mistake!). The only reason why you can't use integrals is because $dn < 1$, and, fun fact, in the past the theories about integrals were using summations, and now you accidentally tried to turn it the other way around. How funny things can be.
