7
$\begingroup$

Given a convergent sequence $x_{n}$ and bounded sequence $y_{n}$ I need to prove that $\limsup (x_{n}+y_{n})=\lim x_{n}+\limsup y_{n}$, when $n$ tends to $\infty$.

I chose $z_{n}=x_{n}+y_{n}$, we know that $z_{n}$ is bounded as being sum of two bounded sequences, so from the Bolzano-Weierstrass Theorem, we know that there is a subsequence of $z_{n}$, let's call it $z_{n_{k}}$, that converges to $\limsup (x_{n}+y_{n})$. $y_{n_{k}}$ is bounded as well, so there is a convergent subsequence $y_{n_{k_{j}}}$. All this gives me that $\limsup (x_{n}+y_{n})\leq \lim x_{n}+\limsup y_{n}$,

What can I do for getting the other inequality?

Thank you and Good evening.

$\endgroup$
  • 1
    $\begingroup$ See also math.stackexchange.com/q/8489 $\endgroup$ – Jonas Meyer Apr 19 '11 at 20:51
  • $\begingroup$ Perhaps it is woth noticing that you have in fact showed $\limsup(x_n+y_n)\le \limsup x_n + \limsup y_n$ for any two sequences (even without the assumption that one of them converges). This inequality is sometimes useful. $\endgroup$ – Martin Sleziak Jul 20 '11 at 10:02
5
$\begingroup$

Simply take a convergent subsequence $y_{n_k}$ such that $\limsup y_n=\lim y_{n_k}$. Then $\limsup(x_n+y_n) \geq \lim (x_{n_k}+y_{n_k})= \lim x_n+\limsup y_n$

$\endgroup$
0
$\begingroup$

Hint:

Suppose $\{y_{n_k}\}$ is a convergent sub-sequence of $\{y_n\}$. What is the limit of $\{z_{n_k}\}$?

$\endgroup$

Your Answer

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