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I tried to prove the following theorem :

If $n>1$ then $\displaystyle1+\frac{1}{2}+\dots+\frac{1}{n} = \frac{k}{m}$ where $k$ is odd and $m$ is even.

and I'd like to know if there is any flaw in my argument.

Proof :

For $n = 2$ and $n=3%$ we verify the theorem directly.

When $n \geq 4$ we can argue as follows :

First of all we sum up all the fractions obtaining

$$1+\frac{1}{2}+\dots+\frac{1}{n}= \frac{n!+1\cdot3\cdots n+\dots+(n-1)!}{n!} \:\:\:\:\:\:\:(1)$$

Now I define the multi-parity function $P$ as a function which counts the number of twos in the factorization of an integer, for example $44 = 2^2 \cdot 11$ so $P(44) = 2$.

Every product $c$ in the numerator of $(1)$ has $P(c) \geq P(n!) - \max[P(x)]$ where $1\leq x \leq n$ . This is true because every product in the numerator has just one factor less than $n!$ , for example we know that $P(1\cdot 3 \cdot ... \cdot n)=P(n!)-1$.

Moreover we know that $P(c) = P(n!) - \max[P(x)]$ in only one case that is when $c$ hasn't the greatest power of two less than $n$ in the product.

So using $g = \max[P(x)]$, which is also always equal or greater than $1$, and $q = P(n!) - g$ we can gather $2^q$ in the numerator and $2^{q+g}$ in the denominator : $$\frac{n!+1\cdot3 \cdots n+\dots+(n-1)!}{n!} = \frac{2^q(a_1+\dots+a_n)}{2^{q+g}m} = \frac{a_1+\dots+a_n}{2^{g}m}$$

but the denominator is even because $g \geq 1$ and the numerator is odd since all the $a_i$ are even except for one.

Is my argument sound ?

Anyway I found the problem in the great book Introduction to algorithms by Udi Manber and I found there are lots of threads about this problem on MSE and my proof is similar to the elementary proof of the statement without Bertrand Postulate but I enjoyed proving it by myself :-)

Clyde Kertzer
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Tortar
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1 Answers1

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Your argument seems sounds, but it's pretty involved. Here's a simpler proof in case you're interested.

Suppose the statement is true for $2,3,\ldots,n - 1$.

Consider:

$$\frac a c = 1 + \frac 1 3 + \cdots + \frac 1 p$$ where $p$ is the largest odd number less than $n$ and

$$\frac b d = \frac 1 2 + \cdots + \frac 1 q$$ where $q$ is the largest even number less than $n$.

Then we have that

$$\frac b d = \frac 1 2 + \cdots + \frac 1 q = \frac 1 2 \left(1 + \frac 1 2 +\frac 1 3+ \cdots + {1\over{q/2}}\right)$$

so $b$ is odd and $d$ is even.

When $q = 2$, $b = 1$, and $d = 2$.

By induction hypothesis we know that for $q > 2$, we have $$1 + \frac 1 2 + \frac 1 3 + \cdots + {1\over{q/2}}$$ so we see that $d$ is even and $b$ is odd.

Now we know $c$ must be odd because $1,3,5,\ldots,p$ are all odd.

Lastly,

\begin{align} 1 + \frac1 2 + \frac 1 3 + \cdots + \frac 1 n &= \frac a c + \frac b d\\ &={ad + bc \over cd} \end{align} so we see that the numerator is odd and the denominator is even.

Clyde Kertzer
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