3
$\begingroup$

Consider $\displaystyle\sum_{1}^{p-1}\frac{1}{x}$, where $p$ is an odd prime. I want to find a general formula for this sum.

I can't believe that I am having trouble figuring this out, but I can't figure out the summation in the question. I have to find a general formula $A_p/B_p$ for the summation from 1 to $p-1$ of $\frac{1}{x}$. Then I have to prove that this is correct and make a conjecture about $A_p \mod p^2$. Somehow, I am having trouble on the first step. Is there a general formula for the summation of $\frac{1}{x}$ alone? That would probably give me a hint where to look. Thanks.

  • 0
    Thanks to Fabian for fixing my incompetence with the symbols and special characters.2011-03-09
  • 0
    Try using a common denominator.2011-03-09
  • 0
    Yeah, I've been working on that. My common denominator would have to be (p-1)!, but then I'm left with gook: (p-1)!/(p-1)! + (p-1)!/2(p-1)! + ... + (p-1)!/(p-1)*(p-1)!2011-03-09
  • 2
    You won't get any better than that. Note also that you can use a better common denominator - not sure what they meant you to do.2011-03-09
  • 0
    What better common denominator, if you don't mind sharing?2011-03-09
  • 0
    Use the best common denominator. Say when adding $1/4 + 1/6$, would you rather use $24$ or $12$?2011-03-09
  • 0
    @Ross: Please try to make the body of your messages self-contained, not relying on the subject/title for key information. The subject is meant to be like an index, or the title of a book on the spine.2011-03-09
  • 0
    @Yuval: The least common multiple. But if I'm trying to get a general formula for this, wouldn't using Euclid's algorithm to find the lcm complicate things? I would probably have to come up with a recursive formula so that I could apply it to all p-1 cases.2011-03-09
  • 0
    The conjecture that you need to make should be obvious if you just compute $A_p/B_p$ for the first few odd primes.2011-03-09
  • 0
    Take a look at http://en.wikipedia.org/wiki/Wolstenholme%27s_theorem. Watch out: spoiler!2011-03-09

1 Answers 1

5

Here's a bigger hint. Try pairing terms in the sum like so:

$$\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{p-1} = \left(\frac{1}{1} + \frac{1}{p-1}\right) + \left(\frac{1}{2} + \frac{1}{p-2}\right) + \cdots + \left(\frac{1}{\frac{p-1}{2}} + \frac{1}{p - \frac{p-1}{2}}\right).$$

Now, get a common denominator for each pair. That will tell you something about $A_p \mod p$. From there maybe you can make some conjectures about $A_p \mod p^2$.

I guess I'm also a little confused as to what you're being asked to do with finding a formula for $\frac{A_p}{B_p}$. This is called the $p$th harmonic number $H_p$, and there is no known nice formula for $H_p$ or for $A_p$ when the fraction is in reduced form. I'm not even sure there is a nice formula for $B_p$ when the fraction is in reduced form. So I think Yuval Filmus's comment is right: You're not going to be able to do much better than the "gook" you are describing in your comment above.

  • 0
    Oh. That explains why there is a warning in my textbook: this question is really hard. Haha.2011-03-09