Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $n=p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$ is the prime factorization of $n>1$ then show that :

$$1>\frac{n}{ \sigma (n)} > \left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\cdots\cdots\left(1-\frac{1}{p_r}\right)$$

I have solved the $1^\text{st}$ inequality($1>\frac{n}{ \sigma (n)}$) and tried some manipulations on the right hand side of the $2^\text{nd}$ inequality but can't get much further.Please help.

share|improve this question
1  
Did you mean $1-\frac1{p_2}$ instead of $1-\frac1{p_2}$? BTW if both sides are multiplicative functions, it suffices to prove the inequality for $n=p^a$, i.e. for powers of primes. –  Martin Sleziak Jun 12 '12 at 17:41
    
@Chris oh sorry I have corrected –  Saurabh Jun 12 '12 at 17:41
add comment

2 Answers

up vote 3 down vote accepted

Note that the function $\dfrac{n}{\sigma(n)}$ is multiplicative. Hence, if $n = p_1^{k_1}p_2^{k_2} \ldots p_m^{k_m}$, then we have that $$\dfrac{n}{\sigma(n)} = \dfrac{p_1^{k_1}}{\sigma \left(p_1^{k_1} \right)} \dfrac{p_2^{k_2}}{\sigma \left(p_2^{k_2} \right)} \ldots \dfrac{p_m^{k_m}}{\sigma \left(p_m^{k_m} \right)}$$ Hence, it suffices to prove it for $n = p^k$ where $p$ is a prime and $k \in \mathbb{Z}^+$.

Let $n=p^k$, then $\sigma(n) = p^{k+1}-1$. This gives us that $$\dfrac{n}{\sigma(n)} = p^k \times \dfrac{p-1}{p^{k+1}-1} = \dfrac{p^{k+1} - p^k}{p^{k+1} - 1} = 1 - \dfrac{p^k-1}{p^{k+1}-1}.$$ Since $p > 1$, we have that $p(p^k-1) < p^{k+1} - 1 \implies \dfrac{p^k-1}{p^{k+1}-1} < \dfrac1p \implies 1 - \dfrac{p^k-1}{p^{k+1}-1} > 1 - \dfrac1p$. Hence, if $n=p^k$, then $$\dfrac{n}{\sigma(n)} > \left( 1 - \dfrac1p\right)$$ Since, $\dfrac{n}{\sigma(n)}$ is multiplicative, we have the desired result.

share|improve this answer
add comment

Let $S$ be the set of products of powers of the prime divisors of $n$ and $n/S$ the set of ratios $(n/s) : s \in S$.

$\frac{n}{\Pi (1 - \frac {1}{p_i})} > \sigma(n) \quad $ is the statement that the sum of all rational numbers in $n/S$ is larger than the sum of all integers in the same set.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.