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 Apr1 asked Finding the limit of $\sqrt[n]{{kn \choose n}}$ Mar30 awarded Critic Mar23 comment Finding $\sup$ of $\{\sqrt{n} - \lfloor\sqrt{n}\rfloor | n\in\mathbb{N}\}$ hmm I understand that, still not sure how to use it though :P Mar23 comment Finding $\sup$ of $\{\sqrt{n} - \lfloor\sqrt{n}\rfloor | n\in\mathbb{N}\}$ Well that's a really nice way of doing it. Didn't think about breaking the number into easier to manage numbers >< Mar23 accepted Finding $\sup$ of $\{\sqrt{n} - \lfloor\sqrt{n}\rfloor | n\in\mathbb{N}\}$ Mar23 comment Finding $\sup$ of $\{\sqrt{n} - \lfloor\sqrt{n}\rfloor | n\in\mathbb{N}\}$ We haven't actually learned limits yet so I'm not sure how to use this, but it seems to be going to 1? no idea how to prove it though. Mar23 asked Finding $\sup$ of $\{\sqrt{n} - \lfloor\sqrt{n}\rfloor | n\in\mathbb{N}\}$ Mar16 comment Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$ Another question, if you don't mind. What was the sign that the original inequality wasn't strong enough for the induction? Is me getting stuck there it or was there something else I need to notice? Mar16 accepted Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$ Mar16 comment Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$ hmm, all clear now, though I'm still pretty sure I never would have figured it out myself. Thank you for the help! Mar16 comment Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$ Oh yeah I see it, I imagined the reversed order :P. Thank you very much! Mar16 comment Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$ I don't understand some of the steps actually, mainly the first (adding the square root of j to the denominator) and the last (going from the sum to n), but it looks pretty neat :) Mar16 comment Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$ I'm looking at this and it seems pretty much like what I did, though obviously with \sqrt{n+1} instead of \sqrt{n}, but your induction step gave the elements \sqrt{k+1}-\sqrt{k} in the reverse order compared to mine. Even knowing that I still can't find what I did wrong... Mar16 asked Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$ Mar16 revised Generalized Bernoulli's inequality of the form $\frac{1}{1-\sum_{i=1}^nx_i}\geq\prod_{i=1}^n(1+x_i)$ added 472 characters in body Mar15 asked Generalized Bernoulli's inequality of the form $\frac{1}{1-\sum_{i=1}^nx_i}\geq\prod_{i=1}^n(1+x_i)$ Jul2 awarded Curious Apr18 awarded Nice Question Nov2 awarded Yearling Dec23 comment limits calculus This is about the same material where I'm at and I found the following to be a great source, examples are much better and more diversified than those given in class and you can really understand how they got to the solutions. math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/…