Sum of divisors

**miz.perfect84** · June 13th 2009, 07:58 AM

[FONT=Times New Roman]I need to prove that for all positive integers n, N^2 is >=2

**TheAbstractionist** · June 13th 2009, 10:12 AM

Originally Posted by miz.perfect84

I need to prove that for all positive integers n, σ (2n) > 2σ (n), where σ (n) denotes the sum of the divisors of n. [Hint: put $n = \color{red}2^r\color{black}m$ , where r > = 0 and m is odd, and consider the cases r = 0 and r >0.]

Use the fact that if $\gcd(m,n)=1,$ then $\sigma(mn)=\sigma(m)\sigma(n).$ If $n$ is odd, then $\sigma(2n)=\sigma(2)\sigma(n)=3\sigma(n)>2\sigma(n ).$ Otherwise, $n=2^rm$ where $r\ge1$ and $m$ is odd, so

$\sigma(2n)\ =\ \sigma(2^{r+1})\sigma(m)$

$=\ (1+2+2^2+\cdots+2^{r+1})\sigma(m)$