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$$\sum^N_{n=1}\liminf_{k \to \infty} f_k(n) = \lim_{k \to \infty} \sum_{n=1}^N \inf_{j \ge k} f_j(n)$$

I am not sure that equation true. Is that equation true? Then why is it?

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Why don't you specify what are those objects you're working with? –  Marra May 8 '12 at 20:23
    
I just wanna solve math.stackexchange.com/q/142728/30883 problem. But I have no basic knowledge wanna have. –  japee May 8 '12 at 20:41

1 Answer 1

up vote 3 down vote accepted

It is known that for any sequence $\{a_k:k\in\mathbb{N}\}\subset\mathbb{R}$ we have $$ \liminf\limits_{k\to\infty}a_k=\lim\limits_{k\to\infty}\inf\limits_{j\geq k}a_j $$ so $$ \sum\limits_{n=1}^N\liminf\limits_{k\to\infty}f_k(n)= \sum\limits_{n=1}^N\lim\limits_{k\to\infty}\inf\limits_{j\geq k}f_j(n)= \lim\limits_{k\to\infty}\sum\limits_{n=1}^N\inf\limits_{j\geq k}f_j(n) $$

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Oh thanks. Is last equation undoubtfully true?(switching sum and limit) –  japee May 8 '12 at 20:46
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It is true, because sum is finite and doesn't depend on limit's variable. –  no identity May 8 '12 at 20:54
    
Oh I got it. Thanks @norbert! –  japee May 8 '12 at 21:14
    
@japee Not at all! –  no identity May 8 '12 at 21:30
    
@japee By the way, you can accept this answer and answers for other your questions - click grey checkmark under two vertical arrows –  no identity May 8 '12 at 21:35

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