In this recent post: Verification of a proof for a classic problem , I tried to prove that the $n$th harmonic number, $H_n$ (the sum of the reciprocals of the first $n$ integers), is never an integer. However, as pointed out by other users, there was a critical flaw in the last step of my "proof", which rendered the whole thing invalid. Hence, after thinking about it for a day, I believe I have come up with a new proof, again by contradiction, that should (hopefully) be valid:

Clearly, when $n=2$, $1+\frac{1}{2}=\frac{3}{2} \notin \mathbb{Z^+}$, and when $n=3$, $1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6} \notin \mathbb{Z^+}$.

Now, suppose that: $$\exists \ k \in \mathbb{Z^+}, k \ge 4, s.t. 1+\frac{1}{2}+...+\frac{1}{k}=q, q \in \mathbb{Z^+}.$$

Our intention is to multiply both sides of the equation by a suitably large integer such that both the L.H.S. and the R.H.S. are integers, then derive a contradiction by analysing parity. Indeed, the most intuitive thing to do is to attempt $k!$ as a multiplier, which was kind of what I did in my previous post, but this will go no-where since we will end up having both the L.H.S. and the R.H.S. being even integers.

Enough of the digression. Let us first define $N$ as the product of all the odd integers from $1$ to $k$. Next, let us define $m$ as the largest positive integer, such that $2^m \le k$. Now, multiply both sides of the equation by $2^mN$. The R.H.S., being $2^mNq$, is clearly an even integer. The L.H.S. is slightly trickier to analyse, but still pretty much doable. Consider each term of the form $\frac{2^mN}{l}, l \in \{1,2,...,k\}$:

Case $1$: $l$ is odd. Then $\frac{2^mN}{l}$ is clearly even, since $l \mid N$.

Case $2$: $l=2^m$. Then, $\frac{2^mN}{2^m}=N$ is odd, by definition of $N$.

Case $3$: $l$ is even and $l\ne 2^m$. Then, we argue that the largest power of $2$ dividing $l$ must be strictly smaller than $m$. Otherwise, if $l=n \cdot 2^m, n \ne 1$, we have $l=n \cdot 2^m \ge 2 \cdot 2^m = 2^{m+1} > k$, by definition, and clearly this is a contradiction! Hence, more precisely, we may write $l$ as $l=2^rp$, where $r < m$, $2^r \mid\mid l$, and $p$ is any odd number $\ge 1$. Thus we have $\frac{2^mN}{l}=\frac{2^mN}{2^rp}$, which is clearly even, since $2^r \mid 2^m $ and $p \mid N$.

Combining the above $3$ cases, we conclude that L.H.S. $=$

$$ \sum_{l=1}^k \frac{2^mN}{l} = \sum_{l \ne 2^m} \frac{2^mN}{l} + N$$ is clearly odd.

But since the L.H.S. is an odd integer, and the R.H.S. is an even integer, we obtain our desired contradiction!

How is the presentation of this proof? Is there any major flaws that I have overlooked? Or is there any portions of it that require more elaboration?


1 Answer 1


I don't see anything wrong with this proof. Note I've seen this sort of parity or power of $2$ contradiction type method used in various ways to prove a particular sum of fractions is not an integer, including for harmonic numbers. For example, there's the answer in Verify my elementary proof that the nth harmonic number is never an integer and most (including the top $2$ scored ones) of the answers in Is there an elementary proof that $\sum \limits_{k=1}^n \frac1k$ is never an integer?.

Regarding your proof, I have a few relatively minor suggestions. First, you didn't need to explicitly mention the first $2$ cases of $n = 2$ and $n = 3$. All you need for your proof to work is that with $2^m \le k$ you have $m \ge 1$, i.e., $k \ge 2$.

Another fairly minor suggestion is you perhaps could have explicitly explained the LHS is an integer when you multiplied by $2^mN$. This is because each denominator is $2^jq$ for some integer $j \le m$ and odd integer $q \le k$, so $2^jq \mid 2^mN$.

One other minor point is you only really needed $2$ cases, not $3$. Your first case could have been included with your third one since it deals with those values with a power of $2$ less than $m$, including a power of $0$, i.e., odd integers. If you did this, the first sentence could have then just been "$l\ne 2^m$", with no other changes to what you wrote. This is possible because in your statement of

Hence, more precisely, we may write $l$ as $l=2^rp$, where $r < m$, $2^r \mid\mid l$, and $p$ is any odd number $\ge 1$.

you would have $r = 0$ for odd values of $l$. Nonetheless, one final very minor point is that $2^r \mid\mid l$ is not needed since $p$ being odd implies this.

These are all minor, basically nit-picking points, but I thought you may appreciate feedback on them.

  • $\begingroup$ Thank you! Will keep these suggestions in mind. $\endgroup$
    – Bananas
    Jul 31, 2020 at 5:21

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