The problem was to show that
$$ H_n = \sum_{k=1}^{n} \frac{1}{k} = 1 + \frac{1}{2} + \dots + \frac{1}{n}$$
is never an integer for $ n \gt 1$.
Solution
There are multiple solutions. One of them uses Bertrand's postulate, which says that there is a prime between $n$ and $2n$ for any $n \gt 1$.
Now consider the largest prime $p$ in $1, 2, \dots, n$.
The denominator of $H_n$ is divisible by $p$. The numerator has terms divisible by $p$, plus the term $1 \times 2 \times \dots \times (p-1) \times (p+1) \times \dots \times n$
By Bertrand's postulate, we have that $2p \gt n$ (otherwise $p$ won't be the largest prime).
Thus the numerator is not divisible by $p$. Thus $H_n$ is not an integer.
$$ H_n = \sum_{k=1}^{n} \frac{1}{k} = 1 + \frac{1}{2} + \dots + \frac{1}{n}$$
is never an integer for $ n \gt 1$.
Solution
There are multiple solutions. One of them uses Bertrand's postulate, which says that there is a prime between $n$ and $2n$ for any $n \gt 1$.
Now consider the largest prime $p$ in $1, 2, \dots, n$.
The denominator of $H_n$ is divisible by $p$. The numerator has terms divisible by $p$, plus the term $1 \times 2 \times \dots \times (p-1) \times (p+1) \times \dots \times n$
By Bertrand's postulate, we have that $2p \gt n$ (otherwise $p$ won't be the largest prime).
Thus the numerator is not divisible by $p$. Thus $H_n$ is not an integer.
good one
ReplyDeleteYeah this one is a quick one (assuming you know the Bertrand result). There are other proofs which don't require any such "advanced" theorems.
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