0
$\begingroup$

I'm confused by limes superior, sup / supremum, same for lim inf...

First of all, is sup an abbreviation of supremum? $sup\Leftrightarrow supremum$?

I have invented a sequence myself and solved it, and I hope I did it correctly.

$$a_{n}=(-1)^{n}\left(1+\frac{1}{n}\right), n \geq 1$$

So, the supremum of this sequence is $\sup=(-1)^{2}\left(1+\frac{1}{2}\right)= 1\cdot\frac{3}{2}= \frac{3}{2}$

The infimum is $\inf=(-1)^{1}\left(1+\frac{1}{1}\right)= -2$

and now lim sup lim inf:

For an even $n$: $\limsup_{n\rightarrow\infty}\left(a_{n}\right )=\left(1+\lim_{n\rightarrow\infty}(\frac{1}{n})\right)=(1+0)=1$

For an uneven $n$: $\liminf_{n\rightarrow\infty}\left(a_{n}\right )=-\left(1+\lim_{n\rightarrow\infty}(\frac{1}{n})\right)=-(1+0)=-1$

I hope I haven't confused anything here? This is really confusing for me and I hope I did it correctly...

$\endgroup$
  • $\begingroup$ This is all 100% correct. $\endgroup$ – mathguy Jul 30 '16 at 22:35
  • $\begingroup$ Thank you very much, so glad to read that :) $\endgroup$ – cnmesr Jul 30 '16 at 22:39
  • $\begingroup$ An interesting exercise for you: Prove that for any sequence ${a_n}$, $\limsup(a_n) = \inf(b_N)$, where $b_N$ is defined as $b_N = \sup_{n\ge N}(a_n)$. Sometimes that is the definition of $\limsup$; rather, use the definition as the $\sup$ of all limits of all subsequences of $(a_n)$. $\endgroup$ – mathguy Jul 30 '16 at 22:40

Your Answer

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

Browse other questions tagged or ask your own question.