So I want to calculate the supremum and infimum of
$\left\{\ln n\right\}_{n=1}^\infty$
and
$\left\{\left(1+\frac{(-1)^n}{2n}\right)^n\right\}_{n=1}^\infty$
separately.
For $\left\{\ln n\right\}_{n=1}^\infty$ I am thinking that I should use the derivative $\frac{1}{n}$ and when I put increasing values of $n$, it approaches zero. Is infimum 1 then as $\frac{1}{1}=1$? And does it not have a supremum?
For $\left\{\left(1+\frac{(-1)^n}{2n}\right)^n\right\}_{n=1}^\infty$ I am thinking that I need to look at odd and even terms separately with Leibniz criterion. But I am very new to it and I don't exactly know how to approach it, and where I should start.
For 1), one of the definitions of $\log$ is $ \int_{1}^{n}\frac{dx}{x} > \sum_{k=2}^{n} \frac{1}{k} $ which is a diverging Harmonic series, so be comparison $\log $ diverges, hence $\limsup \log n =\infty$.
For 2), if you take $x_n = e^{\log x_n}$ (since $x_n>0$), you get (using Taylor series) that $x_n$ oscilates between $\liminf = e^{-\frac{1}{2}}$ and $\limsup=e^{\frac{1}{2}}$. In this case $\lim$ means that for an arbitrarily $\varepsilon-$neighborhood $x_n$ will visit $L_{\sup}- \varepsilon$ and $L_{\inf}+\varepsilon$ infinite number of times.