I was wondering if someone could show me how to do this proof:

If yn converges to L and xn is a bounded sequence, show that

limsup(xn + yn) = limsup(xn) + L.

Thanks for any help. MK

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- October 29th 2007, 04:26 PMMKLyonlimit superior proof
I was wondering if someone could show me how to do this proof:

If yn converges to L and xn is a bounded sequence, show that

limsup(xn + yn) = limsup(xn) + L.

Thanks for any help. MK - October 29th 2007, 06:05 PMThePerfectHacker
- October 30th 2007, 04:33 PMMKLyon
Could someone help with something similar?

I have that (xn) and (yn) are bounded sequences. How would I show that liminf(xn) + limsup(yn) <= limsup(xn + yn)?

It seems easy enough, but I can't figure it out.

Thanks for any help. - October 30th 2007, 07:56 PMThePerfectHacker
Here is a result that you should know.

**Lemmon:**If are convergent sequences and then where is the limit of and is the limit of .

Now we can prove that . The most important thing here is to understand what means. It means the limit of the superior sequence, i.e. *. Now . Since bounded sequences always have limit superiors it means these superior sequences have limits. So by the lemmon: . Thus, .

*)Example. Say . Then , , , .... So , , .... So in general the -th term in the superior sequence is: , thus the limit is . This means . - October 30th 2007, 08:02 PMMKLyon
Thanks, but how does this relate to liminf(xn) + limsup(yn) <= limsup(xn + yn)?

I'm not sure if you misread what I wrote or if I'm missing something, but I need the liminf(xn) not the limsup(xn).