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We have in real analysis that average of every sequence is convergent to the limit of that sequence. So we assume that $\frac{k}n$ for $k=1,2,\ldots,n$ is a sequence and we want to calculate its average. So it is equal to the sum of $\frac{k}{n^2}$. But they have two limits: $\frac12$ and $1$... But Why?

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    A sequence has countably infinitely many terms. You've not given us any such thing.2012-11-02
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    Dear Amin, Welcome to math.SE. since you are a new user, we wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say what your thoughts on the problem are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Further, it would be better if you could typeset your problem so that it is easy for people to read. Kindly look here http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference for more details on typesetting.2012-11-02
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    The limit of $\frac{k}{n}$ as $n$ increases is $0$. So too is the limit of the average as $n$ increases2012-11-02
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    @Henry Sure about that?2013-06-22
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    @Did - no. I have no idea now why I wrote that then. Perhaps I was hinting that "that average of every sequence is convergent to the limit of that sequence" is not particularly helpful here. In fact $$\frac{1}{n}\sum_{k=1}^n\frac{k}{n} = \frac{n(n+1)}{2n^2}=\frac{1}{2} +\frac{1}{2n}$$ which has an obvious limit as $n$ increases.2013-06-22

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