Denominators of partial sums of the harmonic series I have a number theoretic question. Specifically, I am interested in knowing an elementary proof of the following theorem (I used to know it but unfortunately I forgot and can't find it anywhere):
Given $f(n) = 1 + \frac{1}{2} + \frac{1}{3} ... \frac{1}{n}$ and $2^k \leq n < 2^{k+1}$, the largest power of two that divides the reduced denominator of $f(n)$ (i.e. $b$ where $f(n) = \frac{a}{b}$ and gcd($a$,$b$) $=1$) is $2^k$. 
 A: The denominator of $f(n)$ divides $m=\mathrm{lcm}(1,2,3,\dots,n)$ since 
$$
s=mf(n)=\sum\limits_{j=1}^n\frac mj
$$
is a sum of integers, and $f(n)=\frac{\large s}m$. The least terms denominator will divide $m$.
Since $2^k\le n\lt2^{k+1}$, consider the sum
$$
\sum_{j=1}^{2^k-1}\frac1j
$$
The largest power of $2$ in the denominator will be a factor of $2^{k-1}$ since $\mathrm{lcm}(1,2,3,\dots,2^k-1)$ is not divisible by $2^k$.
For $2^k\lt j\lt 2^{k+1}$, there are no multiples of $2^k$. Thus, the least terms denominator of $f(n)-\frac1{2^k}$ is not divisible by $2^k$.
Thus, the least terms denominator of 
$$
\sum_{j=1}^n\frac1j=\underbrace{f(n)-\frac1{2^k}}_{\begin{array}{c}\text{least terms}\\\text{denominator not}\\\text{divisible by $2^k$}\end{array}}+\frac1{2^k}
$$
is a multiple of $2^k$.

Another Approach
Consider $m=\mathrm{lcm}(1,2,\dots,n)/2$. $2^{k-1}\mid m$, but $2^k\not\mid m$. For $1\le j\le n$, each term of $\frac mj$ is an integer, except $\frac m{2^k}$. Thus,
$$
m\sum_{j=1}^n\frac1j
$$
is the sum of $n-1$ integers and one number which is an integer plus $\frac12$.
A: Here's a hint, Ignore the problem. Think about what happens to the power of two of the resultant number when you add together two rational numbers,$\frac{q}{2^kp}+\frac{s}{2^mr}$ where $p,q,s,r$ are all odd.
Use that to prove the theorem inductively.
