Can the sums of two sequences of reciprocals of consecutive integers be equal? I'm primarily a programmer, so forgive me if I don't know the proper nomenclature or notation.
Last night, an old teacher of mine told me about a question that had caused some noodle-scratching for him:
For any two sequences of consecutive integers, can the sums of their reciprocals be equal?
Now, I gather that these sums are called "harmonic numbers" if the consecutive sequence begins with 1.  But what if it doesn't begin with 1?
How might we go about proving that for any two such sequences, their sums are unequal?
I have written a quick Python script that returns, in the form of a reduced fraction, the sum of any $\sigma(m, k)$ where $m$ is the first number to consider and $k$ is the length of the sequence.  
So, I guess I have two questions:
1) Where can I find out about the current state of research on this question?
2) What are the most likely approaches, or "hooks" that I might grasp onto to arrive at a proof?
(Via the comments, I'll add the following for clarity:)
$\sigma(m,k)$ is $\frac{1}{m} + \frac{1}{m+1}+\dots +\frac{1}{m+k−1}.$ My question is: Can there exist distinct pairs $(m_1,k_1)$ and $(m_2,k_2)$ such that the corresponding sums are equal?
 A: Three points which may or may not help.  The first two are obvious.
First, if there is an example where the two sums are equal and the sequences overlap, then removing the overlap from both will provide another example of equal sums.
Second, the consecutive sequence with larger integers and so smaller reciprocals must have more terms than the other sequence if the sums of the reciprocals are equal.
Third, Mathworld says that for the Harmonic number $H_n$, "the denominator is always divisible by the largest power of 2 less than or equal to $n$, and also by any prime $p$  with $n/2 < p \le n$".  A similar statement about primes therefore applies to those appearing in your sequences (at least providing that the primes are over half the end point), and that severely restricts the possibilities for achieving equality.  Something similar may be true with powers of two, and there may be statements to be made about other small primes.
My guess is that it may be possible to show that the denominators of the reduced forms of the sums cannot be equal in any case where the sums are of similar size and the second point above is met.  
A: 
How might we go about proving that for any two such sequences, their sums are unequal?

You can't. For example, 
$$\frac{1}{25}+\frac{1}{757}+\frac{1}{763309}+\frac{1}{873960180913}+\frac{1}{1527612795642093418846225}$$
is the same as $$\frac{1}{33}+\frac{1}{121}+\frac{1}{363}$$.
One keyword is «egyptian fraction»; google should link to lots of information.
