n-th harmonic number is:
$H_n=\sum_{k=1}^n\frac1k$
is there some $n\neq1$ for which $H_n$ is a natural number?
Or can it be proven that there is no such number?
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1$\begingroup$ There is no such number. You can show it by checking that the denominator of $H_n$ is even but the numerator of $H_n$ is odd, for $n\ge 2$. $\endgroup$– Hanul JeonCommented Feb 2, 2017 at 18:03
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1 Answer
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There are no examples other than $n=1$. This can be seen as a result of Bertrand's Postulate. Moreover, for all $n \geq 2$ the numerator of $H_{n}$ is an odd number while the denominator of $H_{n}$ is an even number.
Here is a fairly elementary proof.
answered Feb 2, 2017 at 18:03
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$\begingroup$ Thanks for answer and the link. Although it will take me some time to understand the proof. (I'm just a hobbyist). $\endgroup$– Vlad K.Commented Feb 2, 2017 at 18:18