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Suppose the application $$f:\mathbb{N}^{*}\rightarrow\mathbb{R}$$

$$n\rightarrow\sum_{k=1}^{n}\frac{1}{k}$$How can I prove that $f(\mathbb{N}^*)\cap\mathbb{N}=\{1\}$?

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marked as duplicate by dtldarek, martini, Stefan4024, Vedran Šego, Dan Rust Oct 17 '13 at 12:00

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  • $\begingroup$ I have an idea but I don't think it's a good kick start I tried to prove firstly that $f(\mathbb{N}^*)\cap\mathbb{N} \subset \{1\}$ and then playing with $x \in$ $\endgroup$ – pourjour Oct 17 '13 at 11:19
  • $\begingroup$ it's totally different there are trying to prove that the sum is never integer but read the question here $\endgroup$ – pourjour Oct 17 '13 at 11:29
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Let $\;r\in\Bbb N\;$ be such that $\;2^r\le n!<2^{r+1}\;$ . We have

$$1+\frac12+\ldots+\frac1n=\frac{ n!+\frac{n!}2+\ldots+\frac{n!}{n!}}{n!}$$

Now, $\;2^r\;$ divides the denominator and all the summands in the numerator above except one, namely $\;\frac{n!}2\;$ , and from here that the expression cannot be an integer number...

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