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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?

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  • $\begingroup$ yes, sorry for some typo $\endgroup$ – Jason Ng Oct 19 '18 at 15:15
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$$(1+1)^p-2=\sum_{r=1}^{p-1}\binom pr$$

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 $\displaystyle\binom{p-1}{r-1}=\prod_{m=1}^{r-1}\dfrac{p-1-(m-1)}{m}\equiv(-1)^{r-1}\pmod p$

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