Considering the set of real numbers:
$$A = \left\{\ln\left(\frac{2n+\sqrt{n}}{2n-\sqrt{n}}\right): n \in \mathbb{N} \right\}.$$
I must prove that $0$ is the greatest lower bound of $A$.
I have completed similar examples to this and know I have to show there exists a sequence converging to $0$ but am struggling to find one and prove that $0$ is the glb for this question.
On
Assuming the continuity and monotonocity of the function $\ln$, we find that as $$ \lim_{n \to \infty} \frac{2n + \sqrt{n} }{2n - \sqrt{n} } = \lim_{ n \to \infty} \frac{ 2 + \frac{1}{ \sqrt{n} } }{ 2 - \frac{1}{\sqrt{n} } } = \frac{ 2 + 0}{ 2 - 0} = 1, $$ so we find that $$ \lim_{n \to \infty} \ln \left( \frac{2n + \sqrt{n} }{2n - \sqrt{n} } \right) = \ln 1 = 0, \tag{0} $$ and, for each $n = 1, 2, 3, \ldots$, as $$ \begin{align} &\qquad \frac{ 2 (n+1) + \sqrt{n+1} }{ 2(n+1) - \sqrt{ n+1} } - \frac{2n + \sqrt{n} }{2n - \sqrt{n} } \\ &= \frac{ \left( 2 (n+1) + \sqrt{n+1} \right)^2 }{ 4 (n+1)^2 - (n+1) } - \frac{ \left( 2n + \sqrt{n} \right)^2 }{ 4n^2 - n } \\ &= \frac{ \left( 2 \sqrt{ n+1 } + 1 \right)^2 }{ 4 (n+1) - 1 } - \frac{ \left( 2 \sqrt{n} + 1 \right)^2 }{ 4n -1 } \\ &= \frac{ 4n + 5 + 4 \sqrt{n+1 } }{ 4n + 3 } - \frac{ 4n + 1 + 4 \sqrt{n} }{ 4n-1 } \\ &= \frac{ (4n-1) \left( 4n + 5 + 4 \sqrt{n+1 } \right) - (4n + 3) \left( 4n + 1 + 4 \sqrt{n} \right) }{ (4n+3) (4n-1) } \\ &= \frac{ \big( (4n-1) ( 4n + 5 ) - (4n+3) ( 4n + 1) \big) + \big( 4(4n-1) \sqrt{ n+1} - 4 (4n+3) \sqrt{n} \big) }{ (4n+3) (4n-1) } \\ &= \frac{ 2 + 16n \left( \sqrt{ n+1} - \sqrt{ n} \right) - 4 \left( \sqrt{ n+1} + 3\sqrt{ n} \right) }{ (4n+3) (4n-1) } \\ &= \frac{ 2 + 4 \left( 4 n \left( \sqrt{ n+1} - \sqrt{ n} \right) - \left( \sqrt{ n+1} + 3\sqrt{ n} \right) \right) }{ (4n+3) (4n-1) } \\ &> \frac{ 4 \left( 4 n \left( \sqrt{ n+1} - \sqrt{ n} \right) - \left( \sqrt{ n+1} + 3\sqrt{ n} \right) \right) }{ (4n+3) (4n-1) } \\ &> 0, \end{align} $$ because $$ \begin{align} &\qquad \left[ 4 n \left( \sqrt{ n+1} - \sqrt{ n} \right) + \left( \sqrt{ n+1} + 3\sqrt{ n} \right) \right] \left[ 4 n \left( \sqrt{ n+1} - \sqrt{ n} \right) - \left( \sqrt{ n+1} + 3\sqrt{ n} \right) \right] \\ &= \left( 4 n \left( \sqrt{ n+1} - \sqrt{ n} \right) \right)^2 - \left( \sqrt{ n+1} + 3\sqrt{ n} \right)^2 \\ &= 16n^2 \left( \sqrt{n+1} - \sqrt{n} \right)^2 - \left( \sqrt{ n+1} + 3\sqrt{ n} \right)^2 \\ &= 16 n^2 \left( 2n + 1 - 2 \sqrt{ n(n+1) } \right) - \left( 4n+1 + 6 \sqrt{ n(n+1) } \right) \\ &= \left( 32 n^3 + 16n^2 - 4n -1 \right) + \left( 32n^2 - 6 \right) \sqrt{ n(n+1) } \\ &= \left( 32 n^3 + 16n^2 - \big( 4n +1 \big) \right) + \left( 32n^2 - 6 \right) \sqrt{ n(n+1) } \\ &> 0, \end{align} $$ and also because $$ \begin{align} 4 n \left( \sqrt{ n+1} - \sqrt{ n} \right) + \left( \sqrt{ n+1} + 3\sqrt{ n} \right) &= (4n+1) \sqrt{ n+1 } - (4n-3) \sqrt{ n } \\ &> (4n-3) \left( \sqrt{n+1} - \sqrt{n } \right) \\ &> 0 \end{align} $$ for each $n = 1, 2, 3, \ldots$.
Thus we have $$ \frac{ 2 (n+1) + \sqrt{n+1} }{ 2(n+1) - \sqrt{ n+1} } > \frac{2n + \sqrt{n} }{2n - \sqrt{n} } $$ for each $n = 1, 2, 3, \ldots$, which implies that $$ \ln \left( \frac{ 2 (n+1) + \sqrt{n+1} }{ 2(n+1) - \sqrt{ n+1} } \right) > \ln \left( \frac{2n + \sqrt{n} }{2n - \sqrt{n} } \right) \tag{1} $$ for each $n = 1, 2, 3, \ldots$.
From (0) and (1) above, we find that, the sequence $$ \left(\ \ln \left( \frac{2n + \sqrt{n} }{2n - \sqrt{n} } \right) \ \right)_{ n \in \mathbb{N} } $$ is a strictly increasing sequence converging to $\ln 1 = 0$. Therefore we can conclude that $$ 0 = \ln 1 = \lim_{n \to \infty} \ln \left( \frac{2n + \sqrt{n} }{2n - \sqrt{n} } \right) = \sup \left\{ \ \ln \left( \frac{2n + \sqrt{n} }{2n - \sqrt{n} } \right) \ \colon \ n \in \mathbb{N} \ \right\}, $$ as required.
Hope this is helpful.
On
From Saaqib Mahmood, for the property of being strictly increasing of the elements in $\mathbb{A}$, I think g.l.b. of $\mathbb{A}$ should be $\log 3$?, I will use $\log$ for the natural logarithm. As we see that $$\log 3=\log\left(\dfrac{2+\sqrt{1}}{2-\sqrt{1}}\right)\le\log\left(\dfrac{2n+\sqrt{n}}{2n-\sqrt{n}}\right)$$ for all $n\ge 1$
Edited: I think he was wrong as $$\left(\dfrac{2n+\sqrt{n}}{2n-\sqrt{n}}\right)=1+\dfrac{2}{2\sqrt{n}-1}$$ which is seen as a decreasing sequence. And from now on, $\displaystyle \lim_{n\rightarrow\infty}\log\left(\dfrac{2n+\sqrt{n}}{2n-\sqrt{n}}\right)=0$, with the decrease in value of the elements, we then have $\inf\mathbb{A}=0$ as desired.
Hint: $$\frac{2n + \sqrt{n}}{2n - \sqrt{n}} = 1 + \frac{2\sqrt{n}}{2n - \sqrt{n}}$$