limit superior proof

**MKLyon** · October 29th 2007, 03:26 PM

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

**ThePerfectHacker** · October 29th 2007, 05:05 PM

Originally Posted by MKLyon

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

$\limsup (x_n+y_n) \leq \limsup x_n + \limsup y_n = \limsup x_n + L$
We will show that,
$\limsup (x_n+y_n) \geq \limsup x_n + \limsup y_n$ .
There exists subsequence $x_{n_k}$ so that $\lim x_{n_k} = \limsup (x_n+y_n)$ (a theorem). Then $\lim y_{n_k} = L$ (because a subsequence of a convergent sequence stays the same). Thus, $x_{n_k}+y_{n_k}$ is a subsequence of $x_n + y_n$ and so $\lim x_{n_k}+y_{n_k}$ is a subsequencial limit so $\lim (x_{n_k}+y_{n_k}) \leq \limsup (x_n + y_n)$ (a theorem - limsup's are largest subsequential limits). Thus, $\limsup x_n + L \leq \limsup (x_n+y_n)$ .

**MKLyon** · October 30th 2007, 03:33 PM

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.

**ThePerfectHacker** · October 30th 2007, 06:56 PM

Originally Posted by MKLyon

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.

Here is a result that you should know.

Lemmon: If $x_n \mbox{ and }y_n$ are convergent sequences and $x_n\leq y_n$ then $x\leq y$ where $x$ is the limit of $x_n$ and $y$ is the limit of $y_n$ .

Now we can prove that $\limsup (x_n+y_n)\leq \limsup (x_n)+\limsup(y_n)$ . The most important thing here is to understand what $\limsup$ means. It means the limit of the superior sequence, i.e. $\limsup (x_n) = \lim (\sup \{x_k| k\geq n\})$ *. Now $\sup\{ x_k + y_k | k\geq n\}\leq \sup\{ x_k|k\geq n\} + \sup\{ x_k|k\geq n\}$ . Since bounded sequences always have limit superiors it means these superior sequences have limits. So by the lemmon: $\lim (\sup\{ x_k+y_k|k\geq n\} ) \leq \lim (\sup\{ x_k |k\geq n\}) + \lim (\sup\{ x_k|k\geq n\} )$ . Thus, $\limsup (x_n+y_n) \leq \limsup (x_n)+\limsup (y_n)$ .

*)Example. Say $x_n = \frac{1}{n}+(-1)^n$ . Then $x_1 = 1 + (-1) = 0$ , $x_2 = \frac{1}{2} + (-1)^2 = \frac{3}{2}$ , $x_3 = \frac{1}{3} + (-1)^3 = - \frac{2}{3}$ , .... So $\sup\{ x_k| k\geq 1\} = \sup \{ x_1,x_2,x_3,... \} = \frac{3}{2}$ , $\sup \{ x_k |k\geq 2\} = \sup \{ x_2,x_3,...\} = \frac{3}{2}$ , $\sup \{x_k | k\geq 3\} = \sup \{x_3,x_4,...\} = \frac{5}{4}$ .... So in general the $n$ -th term in the superior sequence is: $\frac{3}{2},\frac{3}{2},\frac{5}{4},\frac{5}{4},\f rac{7}{6},\frac{7}{6},...$ , thus the limit is $1$ . This means $\limsup \left( \frac{1}{n} + (-1)^n \right) = 1$ .

**MKLyon** · October 30th 2007, 07:02 PM

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).

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