# $\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-…-\frac{1}{(p-1)}$=$\frac{a}{(p-1)!}$. Show that $a \equiv \frac{(2-2^p)}{p}$ (mod p)

In fact, it is a similar question to $$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{(p-1)}$$ However, this question has some changes in the sign. ($$p$$ is an odd prime)

My thought is that I tried $$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{(p-1)}-2\cdot(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{(p-1)})$$ The First half can be written as $$\frac{\sigma_{p-2}}{\sigma_{p-1}}$$. While the second half I am not sure how to convert into a sigma notation since when I multiply $$2$$ into the parenthesis, I am not sure what is the ending.

Any suggestions or methods to continue? Or, there is another better way to do it?

• yes, sorry for some typo – Jason Ng Oct 19 '18 at 15:15

$$(1+1)^p-2=\sum_{r=1}^{p-1}\binom pr$$
Now for $$\displaystyle1\le r\le p-1,\binom pr=\dfrac pr\binom{p-1}{r-1}$$
$$\displaystyle\implies\dfrac{(1+1)^p-2}p\equiv\sum_{r=1}^{p-1}\dfrac1r\binom{p-1}{r-1}\pmod p$$
Finally $$\displaystyle\binom{p-1}{r-1}=\prod_{m=1}^{r-1}\dfrac{p-1-(m-1)}{m}\equiv(-1)^{r-1}\pmod p$$
• my lecturer didn't mention the notation (p r). What does it mean actually? Is it a family of $nCr$? – Jason Ng Oct 19 '18 at 13:10