How do I find the value of $f(2022)$ if $$f(n) = \displaystyle \sum_{x=1}^n \dfrac{\sqrt{x}-x}{\sqrt{x+\sqrt{x}}-x}$$ I tried simplifying and rationalizing the denominator but it is not working here. I am stuck on how to proceed with this. Any help is appreciated.
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2Is it $f(n)$? The way you wrote it doesn't make sense. – Klaus Oct 10 '22 at 07:14
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@Klaus Good remark ! – Jean Marie Oct 10 '22 at 07:46
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@Klaus Rectified. Yes it is $f(n)$, it was a mistake, thanks for pointing out. – Alan Oct 10 '22 at 09:38
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$$\dfrac{\sqrt{x}-x}{\sqrt{x+\sqrt{x}}-x} = \dfrac{(x - \sqrt x)(x + \sqrt{x+\sqrt x})}{x^2 - x -\sqrt x}\= \dfrac{(x - \sqrt x)(x + \sqrt{x+\sqrt x})(x^2 - x +\sqrt x)}{(x^2 - x)^2 - x}$$ But that still leaves a great mess to sum up. – Paul Sinclair Oct 10 '22 at 21:03
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@PaulSinclair This is exactly what I tried but don't know to proceed further from here. – Alan Oct 11 '22 at 05:51
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Sorry - your description sounded to me like maybe you hadn't recognized the second rationalization, so that is why I gave the calculation. But I haven't come up with any ideas on how to do the sum from there either. Maybe there is some way of writing the summand as $a_x - a_{x-1}$, which would make the sum telescope down to $a_n - a_1$ – Paul Sinclair Oct 11 '22 at 11:23
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@PaulSinclair No worries :) I highly appreciate your help, I think my description was indeed cold. I think some veteran problem solver will crack this. – Alan Oct 11 '22 at 16:32
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WolframAlpha gives approximately 2028.3253581873656059828510645836640706909610462359374166313981007. But that doesn't help that much. Where did you find this problem? – F_M_ Oct 15 '22 at 09:50