Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I tried to solve following task, but I am not sure whether my solution is correct. Quest: Find supremum and infimum $$A=\left\{\frac{n - k^2}{n^2 + k^3}:n,k \in \mathbb{N} \right\}$$My attempt:
By substituting some numbers we see that $sup(A)=\frac{1}{5}$ and $inf(A)=-\frac{1}{5}$. Now we need to check it with definition. We need to check if: $$\forall _{n,k \in\mathbb{N}} \frac{1}{5}>\left(\frac{n - k^2}{n^2 + k^3}\right)$$ From this we get $n^2+k^3>5n-5k^2 \Rightarrow n(n-5)>-k^2(k+5)$ what is true, since for all $n>5$ left side is positive and right side is always negative. I think that up to $n=5$ it can be done by hand (is it enough to formally prove the first part of definition of supremum?). Second part of definition requires:$$\forall_{\epsilon>0}\exists_{n_{0}, k_{0}\in \mathbb{N}} \frac{1}{5} -\epsilon<\frac{n_{0} - k_{0}^2}{n_{0}^2 + k_{0}^3}\Rightarrow ...\Rightarrow\epsilon>F(n,k)$$ And from Archimedes we know that it is enough to take $\epsilon>[F(n,k)]+1$.
Is it true? I am not sure, because for supremum $\frac{1}{4}$ also works in definition. (Or the argument of checking manually is enough to eliminate the idea of $sup(A)=\frac{1}{4}$ ?). And the same for infimum. Does it go the same way? Thanks in advance!

share|cite|improve this question
By substituting some numbers we see that... Do we? How? – Did Nov 4 '12 at 12:04
Checking for $n=1, k=i, 0<i<5$ etc.? – fdhd Nov 4 '12 at 12:28
And what about $n\gt1$? Very strange... – Did Nov 4 '12 at 12:31
Why strange? by etc i mean the same for $k=1, n=i, 0<i<5$ – fdhd Nov 4 '12 at 12:32
So you say you checked every (n,k) with n and k from 1 to 4? But plenty of these yield a ratio smaller than -1/5... More importantly, nothing guaratees a priori that other (n,k) would not yield larger or smaller ratios. Whatever. – Did Nov 4 '12 at 12:43
up vote 0 down vote accepted

Denote $f(n,k)=\dfrac{n-k^2}{n^2+k^3}$. When $n,k\in\mathbb{N}$, $$f(1,k)=\frac{1-k^2}{1+k^3}\le \frac{1-k^2}{n^2+k^3}\le f(n,k)\le\frac{n-1}{n^2+k^3}\le\frac{n-1}{n^2+1}=f(n,1).$$ Then it is easy to find $\sup(A)=\dfrac{1}{5}$ and $\inf(A)=-\frac{1}{3}$.

share|cite|improve this answer
How do you find $\frac{1}{5}$? By taking derrivative? Wolphram says, that $max(f(n,1))$ is $\frac{1}{\sqrt{2}} - \frac{1}{2}$? – fdhd Nov 4 '12 at 12:48
@user46034: It is easy to see $f(2,1)=f(3,1)=\dfrac{1}{5}$ but when $n>3$, $f(n,1)<\dfrac{1}{5}$. Your maximum $\dfrac{\sqrt{2}-1}{2}$ cannot be achieved by $n\in\mathbb{N}$. – 23rd Nov 4 '12 at 15:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.