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$A=\left\{\dfrac{1}{n}+\dfrac{1}{n^2} \mathrel{\bigg|} n\in \mathbb N^*\right\}$

I have derived the function and I found $\dfrac{-n(n+2)}{n^4}$, so the function is strictly decreasing.

Then I simply said:

  • to find the maximum value for this function we just need to take the minimum value of the interval which is ($n=1$), so $\sup(A)=\max(A)=2$
  • to find the minimum value for this fuction we need to take the maximum value of the interval which is ($n\rightarrow\infty$), so $\inf(A)=0$

So is my method correct or do I need to demonstrate more?

egreg
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Tébina1
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1 Answers1

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Your method is not completely correct: you can't differentiate a function defined on the natural numbers. But the reasoning can be fixed.

You can consider the function $$ f(x)=\frac{1}{x}+\frac{1}{x^2} $$ defined on $[1,\infty)$; its derivative is $$ f'(x)=-\frac{1}{x^2}-\frac{2}{x^3}<0 $$ so this function is decreasing.

Thus $f(x)\le f(1)$ for every $x\ge1$ and, in particular, $$ \frac{1}{n}+\frac{1}{n^2}\le f(1)=2 $$ for all $n\in\mathbb{N}\setminus\{0\}$. Therefore $\max(A)=2$, because $2\in A$.

Since $\lim_{x\to\infty}f(x)=0$, for each $\varepsilon>0$ there exists $n\in\mathbb{N}\setminus\{0\}$ such that $f(n)<\varepsilon$, which proves that $\inf(A)=0$ as, obviously, $f(n)>0$ for all $n$.

egreg
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