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Assume $\lim_{n\rightarrow\infty}x_{n}$ exists. Prove that for any sequence $y_n$, we have $$\limsup\{x_{n}+y_{n}\} =\lim_{n\rightarrow\infty}x_{n}+\limsup\{y_{n}\}$$

I got stuck on this question while revising and while once again intuitively this makes sense, this is the first time I'm dealing with a proving question like this. Any help would be greatly appreciated

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closed as off-topic by Did, anomaly, Harald Hanche-Olsen, Claude Leibovici, Najib Idrissi Sep 27 '14 at 7:58

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Recall that

1) $S=\limsup s_{n}$ if for any $\epsilon >0$ we have $s_{n}< S+\epsilon$ for all sufficiently large values of $n$ and $s_{n}>S-\epsilon$ for infinitely many values of $n$.

2) $s=\liminf s_{n}$ if for any $\epsilon >0$ we have $s_{n}> s-\epsilon$ for all sufficiently large values of $n$ and $s_{n}<s+\epsilon$ for infinitely many values of $n$.

Here we have $$X=\limsup x_n=\liminf x_n=\lim_{n\to\infty}x_n$$ and $Y=\limsup y_n$. Let $\epsilon>0$ be arbitrary. Then we have $x_n<X+\epsilon, y_n<Y+\epsilon$ for all sufficiently large values of $n$. Hence $x_n+y_n<X+Y+2\epsilon$ for all sufficiently large values of $n$.

Again note that $x_n>X-\epsilon$ for all sufficiently large values of $n$ and $y_n>Y-\epsilon$ for infinitely many values of $n$. Hence $x_n+y_n>X+Y-2\epsilon$ for infinitely many values of $n$. Note that this conclusion follows only because $x_n>X-\epsilon$ for all large $n$. Thus the existence of limit of $x_n$ is essential here. It is now clear that $X+Y=\limsup (x_n+y_n)$.

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