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We are given - $$x_n = \begin{cases} (n+1)/n & \text{if $n$ ids odd}, \cr 0 & \text{ if $n$ is even.}\end{cases}$$

We have to compute lim inf and limsup of the sequence.

In the step, $$\lim_{n \to \infty } x_n = \lim_{n \to \infty}(sup\{x_k: k\ge n\}). $$

Then, the next step after this is -

$$\sup\{x_k : k \ge n\} =\begin{cases} (n+1)/n & \text{if $n$ is odd}, \cr ((n+2)/(n+1)) & \text{if $n$ is even.} \end{cases}$$

I don't understand how they got the equation for if n is even.

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consider $x_n$ where $n$ is odd, we note that this is a decreasing positive subsequence.

Also, $x_n=0$ if $n$ is even, hence the supremum will not take value $0$ due to the presence of odd subsequence which is positive and greater than $1$. The biggest term in $\{ x_k : k \geq n\} $ is $x_j$ where $j$ is the smallest odd number that is at least $n$.

$$\sup \{x_k: k \geq n \}=\begin{cases} x_n & \text{if $n$ is odd}\\ x_{n+1} & \text{if $n$ is even}\end{cases}$$

if $n$ is even, $n+1$ is odd.

$$x_{n+1}=\frac{(n+1)+1}{n+1}=\frac{n+2}{n+1}$$ is the largest terms for the set $\{x_k: k \geq n \}$

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  • $\begingroup$ But, shouldn't the equation be equal to zero(as mentioned in the question)? $\endgroup$ – ChocolateAndMath Nov 29 '16 at 19:25
  • $\begingroup$ Let's consider the case when $n=2$. we are interested in the set $\{ x_2,x_3,x_4, \ldots \}= \{ 0,\frac43, 0, \ldots\}$, we can see that $\frac43$ is certainly bigger than $0$. $\endgroup$ – Siong Thye Goh Nov 29 '16 at 19:28
  • $\begingroup$ Hmm.., but how did we know that we must take this step? How did we know this step will get us closer to our end result? $\endgroup$ – ChocolateAndMath Nov 29 '16 at 22:44
  • $\begingroup$ oh, we are interested in studying limsup, hence that is why from definition we evaluate $\sup \{ x_k" k \geq n\}$ before taking limit. $\endgroup$ – Siong Thye Goh Nov 29 '16 at 22:51

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