Help with "A Simpler Dense Proof regarding the Abundancy Index."
Solution 1:
Hint: Since $q \mid \sigma(2^m)$, then $$q \leq 2^{m+1} - 1,$$ which implies that $$\frac{1}{q} \geq \frac{1}{2^{m+1} - 1}.$$
Can you finish?
Help with "A Simpler Dense Proof regarding the Abundancy Index."
Solution 1:
Hint: Since $q \mid \sigma(2^m)$, then $$q \leq 2^{m+1} - 1,$$ which implies that $$\frac{1}{q} \geq \frac{1}{2^{m+1} - 1}.$$
Can you finish?