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Mar
23
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 ><
Mar
23
accepted Finding $\sup$ of $\{\sqrt{n} - \lfloor\sqrt{n}\rfloor | n\in\mathbb{N}\}$
Mar
23
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.
Mar
23
asked Finding $\sup$ of $\{\sqrt{n} - \lfloor\sqrt{n}\rfloor | n\in\mathbb{N}\}$
Mar
16
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?
Mar
16
accepted Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$
Mar
16
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!
Mar
16
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!
Mar
16
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 :)
Mar
16
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...
Mar
16
asked Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$
Mar
16
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
Mar
15
asked Generalized Bernoulli's inequality of the form $\frac{1}{1-\sum_{i=1}^nx_i}\geq\prod_{i=1}^n(1+x_i)$
Jul
2
awarded  Curious
Apr
18
awarded  Nice Question
Nov
2
awarded  Yearling
Dec
23
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/…
Dec
23
comment Why does the definition of limits of a function have strict inequality?
Yeah about that part I know, every book and class did even explain it (rather then just giving it as is)
Dec
23
accepted Why does the definition of limits of a function have strict inequality?
Dec
23
comment Why does the definition of limits of a function have strict inequality?
I actually just commented asking how exactly would you prove it, since I got into a bit of complications of getting rid of the cases of $|x-a|\leq \delta$ and $|f(x)-a|\leq \epsilon$ in the each direction respectively. This clarifies everything nicely.