Liczby pierwsze + Szereg
: 17 lis 2024, o 10:56
Udowodnić, że \(\displaystyle{ \sum_{m=2}^{\infty} \sum_{p \in P} \frac{1}{p^m} <1 }\).
\(\displaystyle{ p_i}\) to i-ta liczba pierwsza
\(\displaystyle{ \sum_{m=2}^{\infty} \frac{1}{p_i^m} = \frac{\frac{1}{p_i^2} }{1-\frac{1}{p_i} }=\frac{1}{p_i(p_i-1)}=\frac{1}{p_i-1}-\frac{1}{p_i} }\).
\(\displaystyle{ \sum_{m=2}^{\infty} \sum_{p \in P} \frac{1}{p^m} = \sum_{p \in P} \sum_{m=2}^{\infty} \frac{1}{p^m} =
\sum_{p \in P} (\frac{1}{p-1}-\frac{1}{p} )=(\frac{1}{2-1}-\frac{1}{2})+(\frac{1}{3-1}-\frac{1}{3})+(\frac{1}{5-1}-\frac{1}{5}) +...=\\=
\frac{1}{2-1}-(\frac{1}{2}-\frac{1}{3-1})-(\frac{1}{3}-\frac{1}{5-1})-(\frac{1}{5}-\frac{1}{7-1}) +...=1 -(\frac{1}{3}-\frac{1}{4})-(\frac{1}{5}-\frac{1}{6}) +... <1 }\).