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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!

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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

1 Answer 1

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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 $\inf(A)=-\frac{1}{3}$.

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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

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