1. ## Sum of divisors

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

2. 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 $\displaystyle n = \color{red}2^r\color{black}m$, where r > = 0 and m is odd, and consider the cases r = 0 and r >0.]
Use the fact that if $\displaystyle \gcd(m,n)=1,$ then $\displaystyle \sigma(mn)=\sigma(m)\sigma(n).$ If $\displaystyle n$ is odd, then $\displaystyle \sigma(2n)=\sigma(2)\sigma(n)=3\sigma(n)>2\sigma(n ).$ Otherwise, $\displaystyle n=2^rm$ where $\displaystyle r\ge1$ and $\displaystyle m$ is odd, so

$\displaystyle \sigma(2n)\ =\ \sigma(2^{r+1})\sigma(m)$
$\displaystyle =\ (1+2+2^2+\cdots+2^{r+1})\sigma(m)$
$\displaystyle >\ (2+2^2\cdots+2^{r+1})\sigma(m)$
$\displaystyle =\ 2(1+2+\cdots+2^r)\sigma(m)$
$\displaystyle =\ 2\sigma(2^r)\sigma(m)$
$\displaystyle =\ 2\sigma(2^rm)$
$\displaystyle =\ 2\sigma(n)$

3. ## thanks

that made it perfectly clear, thank you