A note on primes dividing alternating sums
We define A_n=\sum_{i=1}^n (-1)^i\frac{1}{i} and we show that, for every prime p, there exists a number n such that A_n\equiv 0 (mod p).
math.GM↗
arXiv subjects
Publications and source records attributed to Antonio M. Oller Marcen.
We define A_n=\sum_{i=1}^n (-1)^i\frac{1}{i} and we show that, for every prime p, there exists a number n such that A_n\equiv 0 (mod p).