Let $\dfrac{1}{a_{k}}=\dfrac{1}{k^2}+\dfrac{1}{k^2+1}+\cdots+\dfrac{1}{(k+1)^2-1}$

I need some ideas to exploit for finding the closed form of $$\sum_{k=1}^{n}\left(\lfloor a_{k}\rfloor +\lfloor a_{k}+\dfrac{1}{2}\rfloor \right)$$ we get $$\dfrac{2k+1}{k^2+2k}<\dfrac{1}{a_{k}}<\dfrac{2k+1}{k^2}$$ Therefore $$\dfrac{k^2}{2k+1}<a_{k}<\dfrac{k^2+2k}{2k+1}$$

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    $\begingroup$ A possibly useful fact (easy to prove): $$\lfloor 2x\rfloor=\lfloor x\rfloor+\lfloor x+\frac12\rfloor.$$ Not sure it helps, but it might. $\endgroup$ – Jyrki Lahtonen Jun 9 '15 at 5:58
  • $\begingroup$ interesting idea,Now we only find $[2a_{k}]$ closed form $\endgroup$ – user246688 Jun 9 '15 at 5:59

Let $$b_{n}=\dfrac{1}{n^2}+\dfrac{1}{n^2+1}+\cdots+\dfrac{1}{n^2+n-1}+\dfrac{1}{n^2+n}+\dfrac{1}{n^2+n+1}+\cdots+\dfrac{1}{n^2+2n}$$ so $$\dfrac{2}{n+1}<\dfrac{1}{n}+\dfrac{1}{n+2}=\dfrac{n+1}{n^2+n}+\dfrac{n}{n^2+2n}<b_{n}<\dfrac{n}{n^2}+\dfrac{n+1}{n^2+n}=\dfrac{2}{n}$$ so we have $$\dfrac{2}{k+1}<\dfrac{1}{a_{k}}<\dfrac{2}{k}$$ so $$k<2a_{k}<k+1$$ so $$\lfloor 2a_{k}\rfloor =k$$ so use indenity $$\lfloor a_{k}\rfloor +\lfloor a_{k}+\dfrac{1}{2}\rfloor=\lfloor 2a_{k}\rfloor=k$$

  • $\begingroup$ Elegant solution! +1 $\endgroup$ – user98186 Dec 31 '15 at 21:01

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