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I am having a problem in proving $X=[1/2+\frac{n}{2n+1}]$

You can see that the $Sup(X)=1$ , but I'm having a problem in proving it.

my attempt: i assumed that there is a smaller Supremum $B=1-e$ $(e>0)$ and tried to show that there is an '$a$' that belongs to $X$ such that $a>B$.

But it does not work.

Thanks for help!

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    Clarify $+ -$.? – 2017-02-17
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    If I take $n=1$ then $X=[1,1]$, assuming I understand $+-$, so that $\sup X=\inf X=1$. You are not providing all details. – 2017-02-17
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    sorry , i fixed it – 2017-02-17

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For the sake of completeness, since the question has been edited in the meantime: (the first version was by the way wrong :)

For $X\equiv X_n= \frac{1}{2}+\frac{n}{2n+1}$ we claim to have $\sup{X}=1$. It is easy to verify, that $1$ is indeed an upper bound of $X$, since: $$ \frac{1}{2}+\frac{n}{2n+1} \leq 1 \quad \Leftrightarrow \quad n\leq \frac{1}{2}(2n+1) \quad \Leftrightarrow \quad 0\leq \frac{1}{2}.$$ Again, suppose $\sup X=B<1$, then there once again exists a number $\epsilon>0$ such that $B=1-\epsilon$. But then we should have for all $n\in \mathbb{N}$ $$ \frac{1}{2}+\frac{n}{2n+1}\leq 1-\epsilon \quad \Leftrightarrow \quad \epsilon \leq -\frac{n}{2n+1}.$$ But the right hand side of the last expression is clearly nonpositive: a contradiction to the fact that $\epsilon>0$.

Hence we have indeed $\sup X =1$. Similar arguments can be used for the infimum (which should be 0.5).