# liminf, limsup, limit

• May 14th 2008, 03:25 PM
TXGirl
liminf, limsup, limit
Hi... I am trying to understand liminf and limsup of sequences.

If I have the sequence

$\displaystyle {\frac{1+k}{1+e^k}} $$\displaystyle {+ sin(k)} what is my liminf and my limsup? The limit itself does not exist, so the two should necessarily be unequal, right? I know that \displaystyle limsup=inf_{k \ge 1} \displaystyle sup_{n \ge k}$$\displaystyle {\frac{1+n}{1+e^n}}$ $\displaystyle +sin(n)$

but how do I determine the sup, if it exists?

Also, for the sequence

$\displaystyle {(2+(-1)^k)^k}$,

am I correct that neither the limit nor the limsup exist but that $\displaystyle {liminf=1}$?

And finally, for the sequence

$\displaystyle {1 + {\frac{(-1)^k}{k^2}}}$

because the $\displaystyle {lim=1}$, that means that $\displaystyle {liminf=limsup=1}$, right?

Many thanks!
• May 14th 2008, 03:56 PM
Plato
Hi TXGirl
You need to learn to use LaTex so that we can read your questions.
I know that is is a real pain, but it is well worth it.
• May 14th 2008, 04:34 PM
TXGirl
Quote:

Originally Posted by Plato
Hi TXGirl
You need to learn to use LaTex so that we can read your questions.
I know that is is a real pain, but it is well worth it.

It's true.... I read a tutorial and changed the posting....

LaTeX Tutorial
• May 15th 2008, 11:05 AM
TXGirl
Quote:

Originally Posted by TXGirl
Hi... I am trying to understand liminf and limsup of sequences.

If I have the sequence

$\displaystyle {\frac{1+k}{1+e^k}} $$\displaystyle {+ sin(k)} what is my liminf and my limsup? The limit itself does not exist, so the two should necessarily be unequal, right? I know that \displaystyle limsup=inf_{k \ge 1} \displaystyle sup_{n \ge k}$$\displaystyle {\frac{1+n}{1+e^n}}$ $\displaystyle +sin(n)$

but how do I determine the sup, if it exists?

Also, for the sequence

$\displaystyle {(2+(-1)^k)^k}$,

am I correct that neither the limit nor the limsup exist but that $\displaystyle {liminf=1}$?

And finally, for the sequence

$\displaystyle {1 + {\frac{(-1)^k}{k^2}}}$

because the $\displaystyle {lim=1}$, that means that $\displaystyle {liminf=limsup=1}$, right?

Many thanks!

Anybody have any ideas?