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Find supremum and infimum of the sequence $$a_n = \frac{2n^2}{4n^2+1},\:\:n\:\in \mathbb{N}.$$

I thought the supremum is $1/2$ and there is no infimum because the sequence gets close to ${0}$, but $0\notin \mathbb{N}$.

In addition, I need to find $n$ that $a_n\in \left(S-0.0080,S\right)$ and $a_n\in \left(I,I+0.0080\right)$.

Brevan Ellefsen
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math112311
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  • If $0 \not \in \mathbb N$ then the infimum (and minimum) is $\frac{2}{5}$ when $n=1$. If $0 \in \mathbb N$ then the infimum (and minimum) is $0$ when $n=0$. – Henry Oct 29 '19 at 09:05
  • Look at $ \dfrac {1}{2+1/2n^2}$ for $n \not =0.$ – Peter Szilas Oct 29 '19 at 09:08

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Well, since the $a_n$ are interpreted as elements of $\mathbb R$ and not of $\mathbb N$ (since they are not even in $\mathbb N$) there will be an infimum. The sequence does not get close to $0$, in fact $$\lim_{n \rightarrow \infty} a_n = \frac{1}{2}.$$ Since the sequence is increasing, the supremum is $1/2$ you will find the infimum by computing the first term.

Levi
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