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For some $x \in \mathbb R$ denote by $\lfloor x \rfloor := \sup\{ k \in \mathbb Z \mid k \le x \}$ the biggest integer smaller than $x$, the so called floor function.

Then in my textbook it is claimed that:

$$ \frac{1}{n} \sum_{i=1}^n \lfloor \frac{n}{i} \rfloor \sim \frac{1}{n} \sum_{i=1}^n \frac{n}{i} $$ where the error as we go from $\lfloor n/i \rfloor$ to $n/i$ is smaller than $1$, hence also in the sum.

Why is the error in the sum smaller than $1$, surely the error between $\lfloor n/i \rfloor$ to $n/i$ is smaller than $1$, but I do not see that this holds true for the error between the sums $\sum_{i=1}^n \lfloor \frac{n}{i} \rfloor$ and $\sum_{i=1}^n \frac{n}{i}$?

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  • $\begingroup$ A more interesting question is how much the two sums differ without the 1/n in front. $\endgroup$ – marty cohen Jan 31 '18 at 14:29
  • $\begingroup$ Thats was what I was trying to determine, as I overlooked the $1/n$, so if you know it, feel free to give an answer! :) $\endgroup$ – StefanH Jan 31 '18 at 14:47
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For the sums (up to $n$) the error becomes $n$, and then you divide the sums by $n$, hence the error is $1$.

I believe you read over the $\frac{1}{n}$ part in the quote.

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  • $\begingroup$ Oh yes, sometimes I am just blind... thank u! $\endgroup$ – StefanH Jan 31 '18 at 14:10

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