# Thread: [SOLVED] Sum of divisors

1. ## [SOLVED] Sum of divisors

Let $\displaystyle k$ denote the number of positive divisors of the positive integer $\displaystyle n$, and $\displaystyle S$ the sum of the positive divisors of $\displaystyle n$. Show that

$\displaystyle k\sqrt{n} < S < \sqrt{2k}n$

2. Originally Posted by Bruno J.
Let $\displaystyle k$ denote the number of positive divisors of the positive integer $\displaystyle n$, and $\displaystyle S$ the sum of the positive divisors of $\displaystyle n$. Show that

$\displaystyle k\sqrt{n} < S < \sqrt{2k}n$
we also need to have $\displaystyle n > 1.$ recall that for any distinct (real) numbers $\displaystyle x_j > 0$ we have $\displaystyle (x_1 + \cdots + x_k)\left( \frac{1}{x_1} + \cdots + \frac{1}{x_k} \right) > k^2.$ thus:

$\displaystyle S^2=\sum_{d \mid n}d \cdot \sum_{d \mid n} \frac{n}{d}=n \sum_{d \mid n} d \cdot \sum_{d \mid n} \frac{1}{d} > nk^2.$

to prove the other inequality first note that since $\displaystyle e^x > 1 + \frac{x^2}{2}$ we have $\displaystyle x > \ln \left(1 + \frac{x^2}{2} \right)$ and so $\displaystyle \sqrt{2k} > \ln(k+1).$ using this we have:

$\displaystyle \frac{1}{n}S=\sum_{d \mid n} \frac{1}{d} \leq \sum_{j=1}^k \frac{1}{j} < \int_1^{k+1} \frac{dx}{x}=\ln(k+1) < \sqrt{2k}.$

3. Great job! (Thanks for specifying $\displaystyle n>1$ ).

Here's how I proved the first inequality; we have $\displaystyle S = \frac{1}{2}\sum_{d|n}\left(d+\frac{n}{d}\right)$, and $\displaystyle \frac{1}{2}\left(d+\frac{n}{d}\right) \geq \sqrt n$ by the arithmetic/geometric mean inequality.

4. Originally Posted by Bruno J.

Here's how I proved the first inequality; we have $\displaystyle S = \frac{1}{2}\sum_{d|n}\left(d+\frac{n}{d}\right)$, and $\displaystyle \frac{1}{2}\left(d+\frac{n}{d}\right) \geq \sqrt n$ by the arithmetic/geometric mean inequality.
oh yeah, this is easier!