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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?

Daniel W. Farlow
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    The upvotes though. – Someone May 05 '15 at 16:26
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    It doesn't appear accurate, See wolfram it appears to give the sequence 0,2,4,6,7,9,11 which is certainly not linear in $n$ – JMoravitz May 05 '15 at 16:29
  • Suddenly all the upvotes are gone... – wythagoras May 05 '15 at 16:30
  • @wythagoras I guess people thought that this question isn't upvote worthy once Wolfram showed identity false :P – Wojowu May 05 '15 at 16:31
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    A more interesting question would be: Can it shown to be 2n-3+o(1)? – wythagoras May 05 '15 at 16:31
  • You can write this as $\frac{n}{\sum_{r=0}^{n}\frac{1}{n}\frac{1}{(1+\frac{r}{n})^2}}$ I wish, this were a integral. I wonder if it shows anything. – Someone May 05 '15 at 16:32
  • @JMoravitz,I think is true $n>2?$ –  May 05 '15 at 16:33
  • What is interesting is that, without the floor function, the asymptotic development of the lhs would be $2 n-\frac{5}{2}+\frac{61}{24 n}+O\left(\left(\frac{1}{n}\right)^2\right)$ (this is related to polylogarithms). – Claude Leibovici May 05 '15 at 16:33
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    @Australia as evidenced by my earlier comment, the statement does not hold for $n=1..4$. It begins to match at $n=5$. For $n=4$ it becomes $\approx 6.01$ before applying the floor function. You say you checked the identity for $n=1..6$, now you say $n=3..6$. Have you actually checked any of the cases at all? – JMoravitz May 05 '15 at 16:42
  • maybe is Calculation wrong by hand,I don't use computer calculation it –  May 05 '15 at 16:44
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    @Wojowu Actually, I think the downvotes are from people who are familiar with OP and the kind of questions often asked--hard questions with little to no context. Some of those questions are fine, but it gets to be annoying/tiresome after a while. – Daniel W. Farlow May 05 '15 at 16:49
  • @MagicMan What I said was intended to be a joke (if you didn't realize) but I can understand what you mean - I can see why this can tend to be tiresome. – Wojowu May 05 '15 at 16:51
  • @Wojowu I was responding more to the "upvote" portion of your comment than the Wolfram remark...Wolfram...what a douche (check out the comments haha). – Daniel W. Farlow May 05 '15 at 16:54

2 Answers2

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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 $n > 4$).

Adhvaitha
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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 $$\dfrac{1}{\displaystyle \sum_{i=n}^{2n} \dfrac{1}{i^2}} < 2n-2 \ \text{for}\ n \ge 5$$

Robert Israel
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