0

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.

PythonDaniel
  • 91
  • 7
  • For the infimum $\ln n$ is an increasing function so the infimum is the first term and the supremum is the limit. For the second sequence, yes, try doing the even and odd terms separately, one is increasing and one is decreasing. Try plotting them. – Qiaochu Yuan Sep 15 '20 at 23:12
  • But is infimum 1 because of the derivative or 0 because ln(1)=0? And then supremum is infinity? – PythonDaniel Sep 15 '20 at 23:19
  • 1
    The derivative doesn’t have anything to do with it except insofar as it tells you $\ln n$ is increasing. For example the infimum of the sequence ${ 0,1,2,3... }$ is also $0$. – Qiaochu Yuan Sep 15 '20 at 23:20
  • Ah I see, I think I finally understand the concept, I just struggled with understanding what I exactly was supposed to look at. – PythonDaniel Sep 15 '20 at 23:30
  • For 2), are you quite sure you need $\sup$, not limit? – Alex Sep 15 '20 at 23:45
  • @Alex the question I have states "What is supremum and infimum of the following series?" – PythonDaniel Sep 15 '20 at 23:48
  • Are you sure that the question used the word series rather than the word sequence. The way that I was taught, given a sequence of #'s, the corresponding series represents the "running subtotals". For example, given the finite sequence ${1,2,3,4}$, I was taught that the supremum of the sequence is $4$, while the supremum of the corresponding series is $10$. – user2661923 Sep 15 '20 at 23:56
  • @user2661923 it might as well be sequence, I am translating from another language so I am not sure D: – PythonDaniel Sep 15 '20 at 23:59
  • What about the symbols used in the original text? I have seen the common convention that $\langle a_n\rangle$ refers to a sequence, while ${a_n}$ refers to a series. However, although my impression is that the former symbolism is fairly standard, I'm not confident at all with my interpretation of the latter symbolism. – user2661923 Sep 16 '20 at 00:04
  • @user2661923 in the original text they use {}, and I don't really know, I managed to solve 1) at least. – PythonDaniel Sep 16 '20 at 00:08
  • Based on your last comment, it's probably not a good idea to try to draw a conclusion based on the symbol(s) used in the original text. – user2661923 Sep 16 '20 at 00:10

1 Answers1

1

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.

Alex
  • 19,262