2

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 $\limsup_{n\to\infty}(a_n+b_n)\geq a+\limsup_{n\to\infty}b_n$.

Martin Sleziak
  • 53,687
blub
  • 4,794

2 Answers2

3

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 $$\limsup (-a_n)=-\liminf a_n=-\lim a_n.$$

VJunior
  • 807
0

$\sup\{a_{n}+b_{n},...\}\geq\sup\{a+b_{n},...\}$ 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.

user284331
  • 55,591
  • 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? – blub Nov 27 '17 at 05:57
  • 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. – user284331 Nov 27 '17 at 06:07
  • 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. – blub Nov 27 '17 at 06:21
  • 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. – user284331 Nov 27 '17 at 06:31