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
    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 explai$n$s why there is a war$n$ing in my textbook: this question is really hard. Haha.2011-03-09