## Partial Sums of Harmonic Series

Consider the partial sums of the harmonic series

$\displaystyle1+\cdots+\frac{1}{N},\indent\forall N\in\mathbb{N}$

These partial sums actually have a special name. The $N^{th}$ partial sum is called the $N^{th}$ harmonic number, denoted by $H_{N}$. Harmonic numbers are interesting for many reasons, one of which is their connection to the Riemann hypothesis. University of Michigan mathematician Jeffrey Lagarias proved that the Riemann hypothesis is equivalent to the inequality

$\displaystyle\sigma(n)\leq H_{n}+\ln(H_{n})e^{H_{n}}$,

for all $n\geq 1$, with the inequality being strict for $n>1$. Here, $\sigma(n)$ denotes the sum of all divisors of $n$.

Today, I am interested in a much more elementary question about the harmonic numbers, which I  came across while perusing Peter Clark’s number theory lecture notes. Obviously, $H_{1}=1$, an integer. But $H_{2}=\frac{3}{2}$, $H_{3}=\frac{11}{6}$, $H_{4}=\frac{25}{24}$… Are there any natural numbers $n>1$ such that $H_{n}$ is an integer? My solution is after the jump.

The answer is no. To prove this, it suffices by the Fundamental Theorem of Arithmetic to show that, for any $N>1$, $H_{N}$ can be written as

$\displaystyle H_{N}=\dfrac{p}{2^{e}q},\indent\text{where }e\geq 1, (p,q)=1,\text{ and }2\nmid(p,q)$

We will prove this claim by induction on $N$. We’ve already established the base case $N=2$ is above, so suppose that $H_{N}=\frac{p}{2^{e}q}$, where $p,q\in\mathbb{N}$, $e\geq 1$, and $2\nmid(p,q)$. Consider

$\displaystyle1+\cdots+\frac{1}{N+1}=\dfrac{p}{2^{e}q}+\frac{1}{N+1}=\dfrac{p(N+1)+2^{e}q}{2^{e}q(N+1)}$

If $N+1$ is even, then we can write $N+1=2^{f}r$, where $2\nmid r\geq1$, to obtain

$\displaystyle\dfrac{p(N+1)+q2^{e}}{2^{e}q(N+1)}=\dfrac{p2^{f}r+q2^{e}}{2^{e+f}qr}$

Without loss of generality, suppose that $f=\min\left\{f,e\right\}$, so that

$\displaystyle\dfrac{2^{f}[pr+2^{e-f}q]}{2^{e+f}qr}=\dfrac{pr+2^{e-f}q}{2^{e}qr}$

But $2\nmid (pr+2^{e-f}q)$, since $pr$ is odd and $2^{e-f}q$ is even. Now suppose that $N+1$ is odd. Then $p(N+1)$ is odd, and therefore $p(N+1)+q2^{e}$ is odd and coprime with $2$.