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

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2 Answers 2

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

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

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  • $\begingroup$ 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, 2017 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, 2017 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, 2017 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, 2017 at 6:31

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