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.


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 $\{x_k: k \geq n \}$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.