My Math Forum Integer part

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 May 13th, 2019, 01:07 AM #11 Newbie   Joined: Sep 2017 From: Belgium Posts: 17 Thanks: 5 And if you want to find the sum of the integer parts instead of the integer part of the sum, there is also this: $\sum\limits_{n\leq x}\lfloor n^{\frac{1}{k}} \rfloor=\lfloor x^{\frac{1}{k}} \rfloor (x+1)-\sum\limits_{n\leq x^{\frac{1}{k}}} n^k$ Thanks from idontknow
 May 13th, 2019, 07:33 AM #12 Global Moderator   Joined: Dec 2006 Posts: 20,623 Thanks: 2076 For 6 terms, one finds $\displaystyle \lfloor y \rfloor = 10 < 1 + \!\int_1^6\! \sqrt{x} dx \approx 10.131$. Thanks from idontknow
 May 13th, 2019, 12:35 PM #13 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,654 Thanks: 2632 Math Focus: Mainly analysis and algebra I can't see how you are excluding 30. This is what I imagine your working to be like (although I'm a bit lost as to where the 2+ comes from). https://www.desmos.com/calculator/jfd5ybw1fz Thanks from idontknow

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