[SOLVED] Sum of divisors

• May 18th 2010, 01:49 PM
Bruno J.
[SOLVED] Sum of divisors
Let $k$ denote the number of positive divisors of the positive integer $n$, and $S$ the sum of the positive divisors of $n$. Show that

$k\sqrt{n} < S < \sqrt{2k}n$
• May 18th 2010, 04:46 PM
NonCommAlg
Quote:

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

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

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

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

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

$\frac{1}{n}S=\sum_{d \mid n} \frac{1}{d} \leq \sum_{j=1}^k \frac{1}{j} < \int_1^{k+1} \frac{dx}{x}=\ln(k+1) < \sqrt{2k}.$
• May 18th 2010, 04:57 PM
Bruno J.
Great job! (Thanks for specifying $n>1$ (Giggle)).

Here's how I proved the first inequality; we have $S = \frac{1}{2}\sum_{d|n}\left(d+\frac{n}{d}\right)$, and $\frac{1}{2}\left(d+\frac{n}{d}\right) \geq \sqrt n$ by the arithmetic/geometric mean inequality.
• May 18th 2010, 05:03 PM
NonCommAlg
Quote:

Originally Posted by Bruno J.

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

oh yeah, this is easier!