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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- Oct 29th 2007, 03:26 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 - Oct 29th 2007, 05:05 PMThePerfectHacker
$\displaystyle \limsup (x_n+y_n) \leq \limsup x_n + \limsup y_n = \limsup x_n + L$

We will show that,

$\displaystyle \limsup (x_n+y_n) \geq \limsup x_n + \limsup y_n$.

There exists subsequence $\displaystyle x_{n_k}$ so that $\displaystyle \lim x_{n_k} = \limsup (x_n+y_n)$ (a theorem). Then $\displaystyle \lim y_{n_k} = L$ (because a subsequence of a convergent sequence stays the same). Thus, $\displaystyle x_{n_k}+y_{n_k}$ is a subsequence of $\displaystyle x_n + y_n$ and so $\displaystyle \lim x_{n_k}+y_{n_k}$ is a subsequencial limit so $\displaystyle \lim (x_{n_k}+y_{n_k}) \leq \limsup (x_n + y_n)$ (a theorem - limsup's are largest subsequential limits). Thus, $\displaystyle \limsup x_n + L \leq \limsup (x_n+y_n)$. - Oct 30th 2007, 03:33 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. - Oct 30th 2007, 06:56 PMThePerfectHacker
Here is a result that you should know.

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

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

*)Example. Say $\displaystyle x_n = \frac{1}{n}+(-1)^n$. Then $\displaystyle x_1 = 1 + (-1) = 0$, $\displaystyle x_2 = \frac{1}{2} + (-1)^2 = \frac{3}{2}$, $\displaystyle x_3 = \frac{1}{3} + (-1)^3 = - \frac{2}{3}$, .... So $\displaystyle \sup\{ x_k| k\geq 1\} = \sup \{ x_1,x_2,x_3,... \} = \frac{3}{2}$, $\displaystyle \sup \{ x_k |k\geq 2\} = \sup \{ x_2,x_3,...\} = \frac{3}{2}$, $\displaystyle \sup \{x_k | k\geq 3\} = \sup \{x_3,x_4,...\} = \frac{5}{4}$ .... So in general the $\displaystyle n$-th term in the superior sequence is: $\displaystyle \frac{3}{2},\frac{3}{2},\frac{5}{4},\frac{5}{4},\f rac{7}{6},\frac{7}{6},...$, thus the limit is $\displaystyle 1$. This means $\displaystyle \limsup \left( \frac{1}{n} + (-1)^n \right) = 1$. - Oct 30th 2007, 07:02 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).