<s>[INST] Find the value of $\left[\frac{1}{\sqrt 2}+\frac{1}{ \sqrt 3}+......+\frac{1}{\sqrt {1000}}\right]$ 
Find the value of $$\left[\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+.....+\frac{1}{\sqrt {1000}}\right]$$.Where [•] denote the greatest integer function.

I am very confused about this problem. I tried to find the upper and lower bound of the function. But I can't find any formulas to find the bounds. Somebody please help me.
 [/INST] This is all about providing an accurate approximation for the involved generalized harmonic sum. I will use a technique (creative telescoping) clearly outlined in the first chapter of these course notes.
We may notice that
$$ \sqrt{n+\frac{1}{2}}-\sqrt{n-\frac{1}{2}} = \frac{1}{\sqrt{n+\frac{1}{2}}+\sqrt{n-\frac{1}{2}}} $$
is a telescopic term and it is, additionally, pretty close to $\frac{1}{2\sqrt{n}}$. In particular
$$ \frac{1}{\sqrt{n}}-2\left(\sqrt{n+\frac{1}{2}}-\sqrt{n-\frac{1}{2}}\right)= d_n\\=-\frac{1}{2\sqrt{n}\left(\sqrt{n+1/2}+\sqrt{n-1/2}\right)^2\left(\sqrt{n+1/2}+\sqrt{n}\right)\left(\sqrt{n-1/2}+\sqrt{n}\right)}$$
is a negative term that behaves like $-\frac{1}{32 n^{5/2}}$ for large $n$s. It follows that
$$ \sum_{k=2}^{1000}\frac{1}{\sqrt{k}} = 2\sum_{k=2}^{1000}\left(\sqrt{k+\frac{1}{2}}-\sqrt{k-\frac{1}{2}}\right)+\sum_{k=2}^{1000}d_k $$
has a distance from
$$ 2\sum_{k=2}^{1000}\left(\sqrt{k+\frac{1}{2}}-\sqrt{k-\frac{1}{2}}\right) = 2\left(\sqrt{1000+\frac{1}{2}}-\sqrt{2-\frac{1}{2}}\right) $$
that is less$^{(*)}$ than $\frac{8}{1000}$. By computing the last quantity it follows that the answer is $\color{red}{60}$.
$(*)$ Proof: we have
$$ |d_k|\leq \frac{1}{48 k^{3/2}}-\frac{1}{48(k+1)^{3/2}} $$
hence
$$ \sum_{k=2}^{1000}|d_k|\leq \sum_{k\geq 2}|d_k|\leq \frac{1}{48\cdot 2^{3/2}}<\frac{8}{1000}.$$
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