<s>[INST] For $n>2, n\in\mathbb{Z}$, why is this true: $\left\lfloor 1/\left(\frac{1}{n^2}+\frac{1}{(n+1)^2}+\cdots+\frac{1}{(n+n)^2}\right)\right\rfloor=2n-3$ Let $n>2$ be a positive integer, prove that
$$\left\lfloor \dfrac{1}{\dfrac{1}{n^2}+\dfrac{1}{(n+1)^2}+\cdots+\dfrac{1}{(n+n)^2}}\right\rfloor=2n-3?$$
before I use hand Calculation $n=2,3,4$,maybe I calculation some wrong,I can't use computer it!
so maybe this result  is right. But how to  prove the statement?
 [/INST] Asymptotically,
$$ \sum_{i=n}^{2n} \dfrac{1}{i^2} = \dfrac{1}{2n} + \dfrac{5}{8n^2} + O(1/n^3)$$
so
$$ \dfrac{1}{\displaystyle\sum_{i=n}^{2n} \dfrac{1}{i^2}} = 2n - \dfrac{5}{2} + O(1/n) $$
Thus your equation will be true for sufficiently large $n$.  With sufficiently good explicit bounds on the $O(1/n^3)$ term, you should be able to prove that it is true for all $n \ge 5$.
EDIT: For an explicit upper bound, since $1/x^2$ is convex, 
$$\dfrac{1}{i^2} \le \int_{i-1/2}^{i+1/2} \dfrac{dt}{t^2}$$
so
$$ \sum_{i=n}^{2n} \dfrac{1}{i^2} \le \int_{n-1/2}^{2n+1/2} \dfrac{dt}{t^2} = 
\dfrac{4(n+1)}{(2n-1)(4n+1)} $$
Call this upper bound $U(n)$.  We have
$$ \dfrac{1}{\sum_{i=n}^{2n} \dfrac{1}{i^2} } \ge \dfrac{1}{U(n)}  = 2 n - \dfrac{5}{2} + \dfrac{9}{4(n+1)} > 2 n - \dfrac{5}{2}$$
For a lower bound,  consider
$$ g(i) = \int_{i-1/2}^{i+1/2} \left( \dfrac{1}{t^2} - \dfrac{1}{4t^4}\right)\; dt
= \dfrac{4 (4 i^2 - 16 i - 1)}{(4 i^2 - 1)^2} $$
Now 
$$ \dfrac{1}{i^2} - g(i) = \dfrac{64 i^3 - 4 i^2 + 1}{(4 i^2-1)^2 i^2} > 0$$
for $i \ge 1$.  Thus
$$ \sum_{i=n}^{2n} \dfrac{1}{i^2} \ge \int_{n-1/2}^{2n+1/2} \left(\dfrac{1}{t^2} - \dfrac{1}{4t^4}\right)\; dt = {\frac {8 \left( n+1 \right)  \left( 96\,{n}^{4}-48\,{n}^{3}-32\,{
n}^{2}+5\,n+1 \right) }{3 \left( 2\,n - 1 \right) ^{3} \left( 4\,n+1
 \right) ^{3}}}
$$
which I will call $L(n)$.  Now 
$$\dfrac{1}{L(n)} - (2n - 2) =  -{\frac {384\,{n}^{5}-1760\,{n}^{4}+584\,{n}^{3}+492\,{n}^{2}-62
\,n-13}{ 8 \left( n+1 \right)  \left( 96\,{n}^{4}-48\,{n}^{3}-32\,{n}^{2
}+5\,n+1 \right) }}
$$
which is negative for $n \ge 5$ (you can verify using Sturm's theorem that the numerator and denominator have no zeros for $n \ge 5$).
Thus $$\dfrac{1}{\displaystyle \sum_{i=n}^{2n} \dfrac{1}{i^2}} < 2n-2 \ \text{for}\ n \ge 5$$
</s>