1
$\begingroup$

I'm trying to prove the following equality regarding the limit superior:

Let $(a_n)_{n\in\mathbb{N}}$ be a convergent sequence and let $(b_n)_{n\in\mathbb{N}}$ be bounded. Then

$\limsup_{n\to\infty}(a_n+b_n)=\lim_{n\to\infty}a_n+\limsup_{n\to\infty}b_n$

I worked out some things but I'm unsure whether they are correct:

I've already prove a theorem stating that $\limsup_{n\to\infty}(a_n+b_n)\leq\limsup_{n\to\infty}a_n+\limsup_{n\to\infty}b_n$ for general $(a_n),(b_n)$. As $(a_n)$ is convergent, the $\limsup_{n\to\infty}a_n$ replaces with a $\lim_{n\to\infty}a_n$ and what remains to show is that $\limsup_{n\to\infty}(a_n+b_n)\geq\limsup_{n\to\infty}a_n+\limsup_{n\to\infty}b_n$ for a convergent $(a_n)$.

Writing the definition of the limit superior, we have that $\limsup_{n\to\infty}(a_n+b_n)=\lim_{n\to\infty}\sup\{a_n+b_n,\dots\}$. Now for this supremum expression, we have that $\sup\{a_n+b_n,\dots\}\geq\sup\{a+b_n,\dots\}$ as $(a_n)$ converges to $a$ and the sequence of suprema should be decreasing.

By the laws of the supremum, we then have $\sup\{a+b_n,\dots\}=a+\sup\{b_n,\dots\}$. Running n to infinity should yield $\lim_{n\to\infty}a+\sup\{b_n,\dots\}=a+\limsup_{n\to\infty}b_n$ and thus $\limsup_{n\to\infty}(a_n+b_n)\geq a+\limsup_{n\to\infty}b_n$.

$\endgroup$
2
$\begingroup$

HINT: Note that $$\limsup b_n=\limsup (b_n+a_n+(-a_n))\leq\limsup (b_n+a_n)+\limsup (-a_n).$$ Now, use that $$\limsup (-a_n)=-\liminf a_n=-\lim a_n.$$

$\endgroup$
0
$\begingroup$

$\sup\{a_{n}+b_{n},...\}\geq\sup\{a+b_{n},...\}$ seems not correct for general, but you are close to the goal. My way: Let $\epsilon>0$, there exists some $N$, for $n\geq N$, $a-\epsilon<a_{n}$, so $a-\epsilon+b_{n}<\sup\{a_{n}+b_{n},...\}$, take $\sup$ to $n\geq N$, then $a-\epsilon+\sup_{n\geq N}\{b_{n}\}\leq\sup\{a_{n}+b_{n},...\}$, but $\limsup_{n} b_{n}\leq\sup_{n\geq N}\{b_{n}\}$, so $a-\epsilon+\limsup_{n}b_{n}\leq\sup\{a_{n}+b_{n},...\}$, $\epsilon>0$ is arbitrarily, the result follows.

$\endgroup$
  • $\begingroup$ As $(a_n)$ gets closer to $a$ and since the sequence of supreme is monotone decreasing, why is the statement of exchanging the nth term with limit not true in general? $\endgroup$ – blub Nov 27 '17 at 5:57
  • $\begingroup$ Actually my question is that, $\sup\{a_{n}+b_{n},...\}$ what does this mean? Does that mean $\sup_{n\geq 1}\{a_{n}+b_{n}\}$? I think if we really want to compare both $\sup\{a_{n}+b_{n},...\}$ and $\sup\{a+b_{n},...\}$ the initial indexes should be indicated clearly, if not, there is a counterexample. $\endgroup$ – user284331 Nov 27 '17 at 6:07
  • $\begingroup$ The supremum i mean is the single supremum from the nth index in the sequence such that for the limsup you consider the sequence of these suprema. $\endgroup$ – blub Nov 27 '17 at 6:21
  • $\begingroup$ Essentially you meant $\sup_{k\geq n}\{a_{k}+b_{k}\}\geq\sup_{k\geq n}\{a+b_{k}\}$ for $n=1,2,...$, but I think this is not true: Let $\{a_{k}\}=\{1,-1,1/2,-1/2,1/3,-1/3,...\}$, $\{b_{k}\}=\{-1,1,-1/2,1/2,-1/3,1/3,...\}$, so $a=0$, and that $\sup_{k\geq n}\{a_{k}+b_{k}\}=0$, $\sup_{k\geq n}\{a+b_{k}\}=1/n$ so $0=\sup_{k\geq n}\{a_{k}+b_{k}\}\geq\sup_{k\geq n}\{a+b_{k}\}=1/n$ is not true. $\endgroup$ – user284331 Nov 27 '17 at 6:31

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.