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This is an exercise from Apostol's number theory book. How does, one prove that $$ \frac{\sigma(n)}{n} < \frac{n}{\varphi(n)} < \frac{\pi^{2}}{6} \frac{\sigma(n)}{n} \quad \text{if} \ n \geq 2$$

I thought of using the formula $$\frac{\varphi(n)}{n} = \prod\limits_{p \mid n} \Bigl(1 - \frac{1}{p}\Bigr)$$ but couldn't get anything further.

Notations:

  • $\sigma(n)$ stands for the sum of divisors

  • $\varphi(n)$ stands for the Euler's Totient Function.

Martin Sleziak
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  • $\sigma$ is the sum of divisors, and $\varphi$ is the totient? – J. M. ain't a mathematician Nov 06 '10 at 12:46
  • @J.M: Yes you are rihgt –  Nov 06 '10 at 12:52
  • $\frac{\pi^2}{6}$ is $\zeta(2)$ so I wouldn't be surprised if the solution involved comparing integrals or sums. I will work in that direction and come back – marwalix Nov 06 '10 at 13:19
  • There is a second part of this problem in Apostol's book, but I can't figure it out. It asks the reader to prove that if $x \geq 2$ then $\Sigma_{n\leq x} \frac{n}{\phi(n)} = O(x)$, where the right hand side of the equation uses the Big-O notation. Any insights? – Jane Roston Nov 22 '11 at 19:19

2 Answers2

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Writing $n=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$ for primes $p_i$ we have

$$\sigma (n) = \prod_{i=1}^k \frac{p_i^{a_i+1} – 1}{p_i – 1}$$

and so

$$\sigma(n)\phi(n) = n^2 \prod_{i=1}^k \left( 1 - \frac{1}{p_i^{a_i+1}} \right) \qquad (1)$$

from which both inequalities follow.

For the RH inequality we note that the expansion of the reciprocal of

$$ \prod_{i=1}^k \left( 1 - \frac{1}{p_i^{a_i+1}} \right)$$

is $< \sum_{r=1}^\infty 1/r^ 2 = \pi^2/6.$

Note that

$$\sigma(n) = \prod_{i=1}^k (1+p_i +p_i^2 + \cdots + p_ i^{a_i}),$$

which is where the formula for $\sigma(n)$ comes from and to obtain $(1)$ we've just multiplied this by

$$\phi(n) = n \prod_{i=1}^k \left( 1 - \frac{1}{p_i} \right),$$

and factored out all the $p_i^{a_i}.$

Derek Jennings
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  • How does reciprocal of $ \prod_{i=1}^k \left( 1 - \frac{1}{p_i^{a_i+1}} \right)$ become $<\frac{\pi^2}{6}$? – Kishalay Sarkar Feb 23 '22 at 12:04
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    @Kishalay Sarkar : Does this help? $\displaystyle\prod_{i=1}^k \left( 1 - \dfrac{1}{p_i^{a_i+1}} \right)^{-1}\leqslant \prod_{i=1}^k \left( 1 - \dfrac{1}{p_i^{2}} \right)^{-1}$$= \displaystyle\prod_{i=1}^{k}\bigg(1+\frac{1}{p_i^2}+\frac{1}{p_i^4}+\cdots\bigg)$$\lt \left(1 + \dfrac1{2^2} + \dfrac1{2^{4}} + \cdots \right) \left(1 + \dfrac1{3^2} + \dfrac1{3^{4}} + \cdots \right) \left(1 + \dfrac1{5^2} + \dfrac1{5^{4}} + \cdots \right) \cdots $$=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{2^4}+\dfrac{1}{5^2}+\dfrac{1}{2^23^2}+\cdots$$ = \displaystyle\sum_{r = 1}^\infty \frac1{r^2}=\frac{\pi^2}{6}$ – mathlove Feb 23 '22 at 17:07
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Is known the infinity production for the Riemann zeta function in the form of $$\zeta(s)=\prod\limits_{\large \forall p\in\mathbb P}\dfrac1{1-p^{-s}}. \qquad(\Re s>1)\tag1$$ Then $$1>\Pi=\prod\limits_{\large p_i}\left(1-p_i^{\large-a_i+1}\right) \ge\prod\limits_{\large p_i}\left(1-p_i^{-2}\right)>\dfrac1{\zeta(2)}=\dfrac6{\pi^2},\qquad(n\ge2).\tag2$$ Using the nice result of Derek Jennings in the form of $$\varphi(n)\,\sigma(n)=n^2\Pi,\tag{$\diamondsuit$}$$ easily to get for $\;n\ge 2:$ $$\color{green}{\mathbf{\dfrac{\sigma(n)}n<\dfrac n{\varphi(n)}<\dfrac{\pi^2}6\,\dfrac{\sigma(n)}n}}.$$

Yuri Negometyanov
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