# Thread: Minimum and Maximum for a convergent sequence

1. ## Minimum and Maximum for a convergent sequence

Let (an) be a convergent sequence of real numbers and let A be the set of all its
terms, i.e.A = {a1, a2, . . . , an, . . . }
Does A has a minimum or A has a maximum (or both) ?
Attempt:
Since the series converges A is bounded .Let s=sup A and i = inf A.
if s>a and e>0=>|an-s|<e=>|an|<s+e or |an|>s-e .....this implies that there exist and ak belonging to A such that ak>s-e.By def s= max{a1......an0}=>an=<s and s belongs to A.
The same goes for i=inf A.
Is this correct?

2. ## Re: Minimum and Maximum for a convergent sequence

Originally Posted by StefanM
Let (an) be a convergent sequence of real numbers and let A be the set of all its terms, i.e.A = {a1, a2, . . . , an, . . . } Does A has a minimum or A has a maximum (or both) ?
(i) If $\displaystyle a_n=1/n$ , $\displaystyle A$ has maximum but not minimum. (ii) If $\displaystyle a_n=-1/n$ , $\displaystyle A$ has minimum but not maximum. (iii) If $\displaystyle a_1=-1,a_n=1/n\;(n\geq2)$ , $\displaystyle A$ has maximum and minimum.

Since the series converges A is bounded .Let s=sup A and i = inf A.
Right, now prove that if $\displaystyle i$ is not a minimum of $\displaystyle A$ and $\displaystyle s$ is not a maximum of $\displaystyle A$ (hence $\displaystyle i<s$ ) there exists a subsequence with limit $\displaystyle i$ and a subsequence with limit $\displaystyle s$ (contradiction) .

3. ## Re: Minimum and Maximum for a convergent sequence

Originally Posted by FernandoRevilla
(i) If $\displaystyle a_n=1/n$ , $\displaystyle A$ has maximum but not minimum. (ii) If $\displaystyle a_n=-1/n$ , $\displaystyle A$ has minimum but not maximum. (iii) If $\displaystyle a_1=-1,a_n=1/n\;(n\geq2)$ , $\displaystyle A$ has maximum and minimum.
And, just to be complete, (iv) if $\displaystyle a_n= 1- 1/n$ for n even, $\displaystyle a_n= 1/n$ for n odd, $\displaystyle A$ has neither max nor min.

Right, now prove that if $\displaystyle i$ is not a minimum of $\displaystyle A$ and $\displaystyle s$ is not a maximum of $\displaystyle A$ (hence $\displaystyle i<s$ ) there exists a subsequence with limit $\displaystyle i$ and a subsequence with limit $\displaystyle s$ (contradiction) .

4. ## Re: Minimum and Maximum for a convergent sequence

Originally Posted by HallsofIvy
And, just to be complete, (iv) if $\displaystyle a_n= 1- 1/n$ for n even, $\displaystyle a_n= 1/n$ for n odd, $\displaystyle A$ has neither max nor min.
Take into account that by hypothesis $\displaystyle a_n$ is convergent.

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### maximum and minimum limits of sequences

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