Let ${x_n}$ be defined by

$$x_n : = \begin{cases} \frac{n+1}{n}, &\text{if } n \text{ is odd}\\ 0,&\text{if } n \text{ is even}. \end{cases}$$

I am pretty sure about $\lim_{n\to\infty}\inf x_n = 0$

because if ${x_1} = 2$, $x_2 = 0$, $x_3 = 4/3$, $x_4 = 0$ so $\lim_{n->\infty}\inf x_n = 0$

But about sup

$$\sup\{x_k : k \geq n\}=\begin{cases} \frac{n+1}{n}, &\text{if } n \text{ is odd}\\ \frac{n+2}{n+1}, &\text{if } n \text{ is even}. \end{cases}$$

I understand about odd but don't understand about when $n$ is even.

Why it is not $0$ when $n$ is even?

  • I would prefer to rewrite $x_n = \begin{cases} 1+ \frac{1}{n}, & n \, \textrm{odd} \\ 0, & n \, \textrm{even} \end{cases}$ – joeb Apr 14 '17 at 0:29
  • yeah I don't know what latex is for that . – Kwangi Yu Apr 14 '17 at 5:06

When $n$ is even, $n+1$ is odd. What is the value of $x_{n+1}$? Is there are ny $x_k$ with $k>n$ and such that $x_k>x_{n+1}$?

  • ah I just got it when n = 2 the set ${x_k : k \geq 2}$ gonna be $x_2 = 0, x_ 3 = 4/3 = \frac{2n+1}{n+1}$ – Kwangi Yu Apr 14 '17 at 0:23
  • so in this case sup gonna be $\frac {2n+1}{n+1}$ Is this right logic? – Kwangi Yu Apr 14 '17 at 0:27
  • It will be $\frac{n+2}{n+1}$. I believe you mistyped that. – Fimpellizieri Apr 14 '17 at 0:30
  • oh yeah I mistype :/ – Kwangi Yu Apr 14 '17 at 1:25

The $\limsup$ of a sequence (when it exists) is defined as the smallest number $X$ such that, for all $\epsilon>0$, there exists an $N_{\epsilon}$ such that, for all $n>N_{\epsilon}$, $x_n < X+\epsilon$. Since there are always terms greater than $1$, no matter how far you go, the $\limsup$ of your sequence cannot be $0$.

  • Yeah I know about that sup cannot be zero, but I did not know that why n is even $\frac{n+2}{n+1}$ – Kwangi Yu Apr 14 '17 at 0:28

You can use the highest exponent limit identity which would get you both limits = 1 on both even and odd cases.

Then, since k is only an index, the limit of the $x_k$s would be the same as the limits of the sequence of $x_n$s, which would be 1. So, the $\sup$ of the set will be $1$, too.

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