# General formula for $\sum_1^{p-1} \frac{1}{x}$, where $p$ is an odd prime

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.

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Thanks to Fabian for fixing my incompetence with the symbols and special characters. –  user7435 Mar 9 '11 at 16:57
Try using a common denominator. –  Yuval Filmus Mar 9 '11 at 16:58
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)! –  user7435 Mar 9 '11 at 17:09
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. –  Yuval Filmus Mar 9 '11 at 17:11
What better common denominator, if you don't mind sharing? –  user7435 Mar 9 '11 at 17:17

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