Proof That 1979 Divides p in Math Series

[oszust001]

Let p,q is natural.
p/q=1-1/2+1/3-1/4+...-1/1318+1/1319.
How can I proof that 1979|p.

 

[BAdhi]

can't we use series for this,

\frac{p}{q}=\sum_{r=0}^{659}\frac{1}{2r+1}-\sum_{r=1}^{659}\frac{1}{2r}

 

[Citan Uzuki]

Here are several hints to get you started. First, observe that1979 is prime, so 1319! is coprime to 1979, so if the 1979 divides 1319!*p/q, then 1979 divides p. Hence you only have to show that the sum of integers 1319!/1 - 1319!/2 + 1319!/3... is divisible by 1979. Some further hints:

Hint 1:

Spoiler

This is equivalent to showing that the sum is congruent to zero mod 1979, so do all your work in the finite field Z/1979Z

Hint 2:

Spoiler

Represent the sum as <tex>1319!\left(\sum_{k=1}^{1319}k^{-1} - 2\sum_{k=1}^{659}(2k)^{-1}\right)</tex> (change the angle brackets to square brackets to see the tex rendered)

Hint 3:

Spoiler

As k goes from 1 to 659, -k goes from 1978 to 1320 (mod 1979)

Hint 4:

Spoiler

If k ranges over the nonzero elements of Z/1979Z, so does k^(-1)