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?
Convergent Of Sum of $\frac{k}{n^2}$
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calculus
real-analysis
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0@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
1 Answers
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$\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{k}{n^2}=\frac12\qquad(\text{and not}\ 1)$