An inequality involving sigma and tau

Let \tau(n) and \sigma(n) respectively be the number and the sum of the positive divisors of the positive integer n. We will prove the following inequality:

\sigma(n)\ge \sqrt n\tau(n).

First proof. By the Cauchy-Schwarz inequality,

\displaystyle\sigma(n)^2=\left(\sum_{d\mid n}d\right)\left(\sum_{d\mid n}\frac nd\right)\ge (\sqrt n\tau(n))^2=n\tau(n)^2.

Second proof. By Chebyshev’s inequality,

\displaystyle\sigma(n)^2=\left(\sum_{d\mid n}d\right)\left(\sum_{d\mid n}\frac nd\right)\ge \tau(n)\cdot n\tau(n).

Third proof. By the AM-GM inequality,

\displaystyle 2\sigma(n)=\sum_{d\mid n}\left(d+\frac nd\right)\ge 2\sqrt n\tau(n).

Fourth proof. Again by the AM-GM inequality,

\displaystyle n^{\tau(n)}=\prod_{d\mid n}d\prod_{d\mid n}n/d=\left(\prod_{d\mid n}d\right)^2\Rightarrow n^{\tau(n)/2}=\prod_{d\mid n}d\le\left(\frac{\sigma(n)}{\tau(n)}\right)^{\tau(n)}.

Fifth proof. If n=p_1^{a_1}\cdots p_k^{a_k} factored into primes, then AM-GM gives

\displaystyle\sigma(n)=\prod_{j=1}^k(p_j^{a_j}+\cdots+p_j+1)\ge\prod_{j=1}^k(a_j+1)p_j^{a_j/2}=\sqrt{n}\tau(n).

We also trivially have \sigma(n)\le n\tau(n). Hence for all n,

\displaystyle \boxed{\sqrt n\le\frac{\sigma(n)}{\tau(n)}\le n}.

In general, if f:\mathbb N\to\mathbb N is completely multiplicative, and

F(n):=\displaystyle\sum_{d\mid n}f(d),

then the first four proofs can be generalised to deduce that

(*)\qquad\qquad\qquad\qquad\quad\ F(n)\ge \tau(n)\sqrt{f(n)}.

For example, if \sigma_k(n) is the sum of the k-th powers of the positive divisors of n, then this shows that

\displaystyle \boxed{n^{k/2}\le\frac{\sigma_k(n)}{\tau(n)}\le n^k}

for each k\ge 0. This, combined with the result from the previous post, gives

\displaystyle\lim_{n\to\infty}\frac{\sigma_k(n)}{n^{k+1}}=0,

i.e., \sigma_k(n)=o(n^{k+1}) as n\to\infty, for each k\ge 0.

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1 Comment

Filed under Number theory

One response to “An inequality involving sigma and tau

  1. Pingback: Another inequality involving sigma and tau | Samin Riasat's Blog

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