Least upper bound of $(-1)^n -(2n+1)^{-\frac{1}{2}}$ 
Find $\text{sup}(A)$ if $A = \left\{(-1)^n-\dfrac{1}{\sqrt{2n+1}}: n \in \mathbb{N^{+}}\right\}.$ Argue straight from the definition.

Let $x_n = (-1)^n-\dfrac{1}{\sqrt{2n+1}}.$ I realised that an upper bound for $x_n$ is $-1-\dfrac{1}{\sqrt{3}}$. I know we want to show that $-1-\dfrac{1}{\sqrt{3}}-\epsilon$ is not an upper bound for $\epsilon > 0$, but I can't find a suitable $\epsilon$. I tried mimicking a solution given to a similar question of mine to no avail; I can't get required inequality.
I said let $\epsilon > 0$, then there exists an odd integer $2n+1$ such that $\dfrac{1}{\sqrt{4n+3}}-\dfrac{1}{\sqrt{3}}< \epsilon$. Which is the same as $ \displaystyle 1+\frac{1}{\sqrt{4n+3}} -\frac{1}{\sqrt{3}} -1 <\epsilon $ which implies $\displaystyle -x_{2n+1} -\frac{1}{\sqrt{3}} -1 <\epsilon$ and this gives me $\displaystyle x_{2n+1} > -\frac{1}{\sqrt{3}}-1-\epsilon$, which makes me think my choice of epsilon is wrong or the lower bound is. 
How do you do this problem straight from the definition of the least upper bound?

To find an upper bound, just consider only the larger terms in $A$. Since $(-1)^n=-1$ for $n$ odd, you can restrict attention to $n$ even for which $(-1)^n=1$. Hence $$\sup(A)=\sup{\left\{1-\frac{1}{\sqrt{2n+1}}:n\in2\Bbb N\right\}}$$ From this point, it should straightforward to find the $\sup(A)$. (just let $n\to +\infty$).

To see that $1$ is the least upper bound, choose $ε>0$ and observe that $1-ε$ is not an upper bound any more for $A$. To do this, just take $n$ big enough, such that $$1-\frac{1}{\sqrt{2n+1}}>1-ε\implies 2n+1>\frac1{ε^2}\implies n>\frac12\left(\frac1{ε^2}-1\right)$$
