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The problem is as follows: Prove that there are infinitely many natural numbers such that $σ(n)>100n$. $σ(n)$ is the sum of all natural divisors of $n$ (e.g. $σ(6)=1+2+3+6=12$).

I have come to the conclusion that if I can prove that there exists at least one such number, then there exist infinitely many of them. Here is how I showed that: Induction - if $σ(n)>100n$, then $σ(pn)>100pn$, where p is a prime number that doesn't divide $n$. Using the multiplicability of the function $σ(x)$ we get: $$σ(pn)=σ(p)σ(n)=(p+1)σ(n)>(p+1)100n>100pn$$ Since there are infinitely many primes the proof of this "lemma" is done. Back to finding one of those n. So the statement, if we use the factorisation of $n$, is as follows: $$(1 + 1/p_1 + 1/(p_1^2) + \dots + 1/(p_1^{a_1})) \dots (1 + 1/p_k + 1/(p_k^2) + \dots + 1/(p_k^{a_k})) > 100$$ So I just need to prove that there exists such primes $p_1, p_2, ..., p_k$, as well as the exponents $a_1, a_2, ..., a_k$.

I have found somewhere that (1 + 1/p_1 + 1/(p_1^2) + ... + 1/(p_1^a_1)) ... (1 + 1/p_k + 1/(p_k^2) + ... + 1/(p_k^a_k)) > 1 + 1/2 + 1/3 + ... 1/m > (anything), but I don't understand why so I am looking for some help.

Bill Dubuque
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Mathology
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1 Answers1

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What you are doing right now:

  1. Prove that if there exists $n\in\mathbb{Z}^+$ such that $\sigma(n)\geq100n$, then there are infinitely many such $n$.
  2. Find such a $n$.

Your step 1 is completely correct, and if your step 2 would work as well. However, step 2 is in fact, quite difficult. Instead, we might just want to show that

  1. There exists such a $n$.

without finding the exact value of $n$. The proof proceeds as follows:

Let $n = \prod_{p_i} p_i ^{a _i}$, where $\{p_i\}$ are distinct primes. It should be clear that $$\sigma(n) = \prod_{p_i} \frac { p_i^{a_i+1} - 1 } { p_i - 1}$$ You can comment under this post if you do not understand this step.

Hence, we want to find $n$ such that $$\frac{\sigma(n)}{n}=\prod_{p_i}\frac { p_i ^{a_i + 1} -1 } { (p_i -1 )p_i ^{a_i} }> 100$$

However, we know that $$a_i \to\infty \implies \frac { p_i ^{a_i + 1} -1 } { (p_i -1 )p_i ^{a_i}} \to\frac { p_i}{(p_i - 1)}$$

But we also have (from here) $$ \prod_{p_i} \frac {p_i}{p_i-1} \to\infty$$

Hence, this guarantees that such a $n$ exists, which completes our proof.

IraeVid
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  • Why is $\prod \frac{p_i}{p_i-1} $ divergent? – David Lui Sep 10 '23 at 08:02
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    https://math.stackexchange.com/questions/4067029/how-to-show-that-prod-p-text-prime-fracpp-1-diverges – IraeVid Sep 10 '23 at 08:14
  • Does anyone know an explicit $n$ such that $\sigma(n)\geq100n$? How large does it need to be? – lhf Sep 10 '23 at 11:46
  • I don't know, but you can try searching for an article or an OEIS sequence. You can also run a computer program if you're interested. However, we do not need to find such $n$ to answer OP's question - a proof of existence is sufficient. – IraeVid Sep 10 '23 at 12:35
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    see https://math.stackexchange.com/questions/349588/how-would-i-find-a-number-where-sum-d-mid-nd-100n where particular $n$ were found. Probably the most useful is one using primorial $n$ – Will Jagy Sep 10 '23 at 15:35
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    Note that you can just assume the $a_i$ are all $1$ and it makes the proof simpler. – Eric Sep 10 '23 at 18:30