Number Theory: Show for all positive integers σ(2n)>2σ(n)

Question

Just need a bit of help with this.

Number Theory: Show for all positive integers σ(2n)>2σ(n)

Sigma being the function of total numbers of factors including 1 and itself from number theory

I know a good starting point would be to consider n=2^r m, but I'm really stuck on how to apply this to achieve the proof.

Thank you.

Answer 1

An alternative solution (again assuming $\sigma(n)$ is the sum of the divisors of $n$):

$\sigma$ is multiplicative. Let $n=2^io$ where $o$ is odd. Then $$\sigma(2n)=\sigma(2^{i+1})\sigma(o) = (2^{i+2}-1)\sigma(o),$$ but $$2\sigma(n)=2\sigma(2^i)\sigma(o)=2(2^{i+1}-1)\sigma(o)=(2^{i+2}-2)\sigma(o).$$

Then we see that $\sigma(2n)=2\sigma(n)+\sigma(o)$, or that $\sigma(2n) > 2\sigma(n)$.

Answer 2

$ \sigma(n)$ is the sum of the divisors of $n$. \begin{eqnarray*} \sigma(n)= \sum_{ d \mid n} d. \end{eqnarray*} Now if $d \mid n $ then $ 2d \mid 2n$ (and there could be other values that divide $2n$) so \begin{eqnarray*} \sigma(2n) \geq \sum_{ 2d \mid 2n} 2d = 2\sum_{ d \mid n} d = 2 \sigma(n). \end{eqnarray*}