# Sum of divisors proof question

• Aug 30th 2010, 01:07 AM
mel240
Sum of divisors proof question
s(n) denotes the sum of divisors function

suppose n=p^k, where p is prime and k is a positive integer
s(p^k)= [(p^k+1)-1]/(p-1)

show:
1+ 1/p < s(n)/n < 1+ 1/(p-1)

Also if p=3 how large does k have to be to make s(n)/n > 1.4999

I know that if the above is expanded then it makes sense that each part is less than the next but I'm not sure if this is a proper proof as I am using what they have given and just expanding it and showing it's true. Is there another way to go about it? and I know k has to be something like 15, but is there an easier way to figure it out rather than just starting from k=1 and working my way up?
• Aug 30th 2010, 09:40 PM
simplependulum
Yes , expanding is a proper proof , or we can set $x = 1/p$ and suppose $n$ is a power of a prime .

Then $\frac{s(n)}{n} = 1+ x + x^2 + ... + x^k$

The LHS from the inequality is $1 + x$ while the RHS is $1 + \frac{x}{1-x} = \frac{1}{1-x} = 1 + x + x^2 + ...$ by combining them we can see what we need to show :

$1+x \leq 1 + x + x^2 + ... + x^k < 1 + x + x^2 + ....$ which is obviously true for $k \geq 1$ .