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.
What you are doing right now:
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
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.