<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] Let $f(n) = \dfrac1{n^2} + \dfrac1{(n+1)^2} + \cdots + \dfrac1{(2n)^2}$.
We then have
$$\int_{n-1}^{2n} \dfrac{dx}{x^2} > \dfrac1{n^2} + \dfrac1{(n+1)^2} + \cdots + \dfrac1{(2n)^2} \geq \int_n^{2n+1} \dfrac{dx}{x^2}$$
This gives us
$$\dfrac1n-\dfrac1{2n+1}  < f(n) < \dfrac1{n-1} - \dfrac1{2n} \implies \dfrac{n+1}{n(2n+1)} < f(n) < \dfrac{n+1}{2n(n-1)}$$
Hence, we have
$$\dfrac{2n^2-2n}{n+1}< \dfrac1{f(n)} < \dfrac{2n^2+n}{n+1} \implies 2n - \dfrac{4n}{n+1} < \dfrac1{f(n)} < 2n - \dfrac{n}{n+1}$$
A tighter bound using this integral approach (ala Euler–Maclaurin) should provide the answer.

EDIT For a more precise answer, we have
\begin{align}
f(n) & = \int_{n^-}^{2n^+} \dfrac{d\lfloor x \rfloor}{x^2} = \left.\dfrac{\lfloor x \rfloor}{x^2} \right \vert_{n^-}^{2n^+} + \int_{n^-}^{2n^+} \dfrac{2\lfloor x \rfloor}{x^3}dx =\dfrac1{2n} - \dfrac{n-1}{n^2} + \int_{n^-}^{2n^+}\dfrac{2x-2\{x\}}{x^3}dx\\
& = \dfrac{n-2n+2}{2n^2} + \dfrac1n - 2\int_{n^-}^{2n^+} \dfrac{\{x\}}{x^3}dx = \dfrac{n+2}{2n^2} - 2\int_{n^-}^{2n^+} \dfrac{\{x\}-1/2}{x^3}dx  -\int_{n^-}^{2n^+} \dfrac{dx}{x^3}\\
& = \dfrac{n+2}{2n^2} - 2\int_{n^-}^{2n^+} \dfrac{\{x\}-1/2}{x^3}dx  -\dfrac3{8n^2} = \dfrac1{2n} + \dfrac5{8n^2} + \mathcal{O}(1/n^3) = \dfrac1{2n}\left(1+\dfrac5{4n} + \mathcal{O}(1/n^2)\right)
\end{align}
This means
$$\dfrac1{f(n)} = \dfrac{2n}{1+\dfrac5{4n} + \mathcal{O}(1/n^2)} = 2n\left(1-\dfrac5{4n} + \mathcal{O}(1/n^2)\right) = 2n-\dfrac52 + \mathcal{O}(1/n)$$
Hence, we have
$$\left\lfloor \dfrac1{f(n)} \right\rfloor = 2n-3$$ eventually (in fact for $n > 4$).
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