Another Needlessly Sophisticated Proof of a Simple Result

This post belongs solely to a genre that should be called "joke proofs". The only thing a joke proof provides (except an awfully sophisticated argument for a very simple result) is intellectual stimulation. Some very well known contributions to this field are a proof of irrationality of $2^{1/n}$ using Fermat's Last Theorem or a proof of infinitude of primes using irrationality of $\zeta (3)$ (although infinitude of primes have received some more suspicious entries as well).

The following theorem (if you can call it a theorem) follows directly from looking at the integers $\{(k+1)!+\ell {\}}_{\ell =2}^{k+1}$. But our proof instead uses two other well known results, namely the fact that the sum of reciprocals of primes diverges (proof) and something slightly stronger than the fact that the sum of reciprocals of twin primes converges (proof). It is well known that an exact same proof technique runs through to show that if ${\mathcal{S}}_k=\{p\in \mathcal{P}:p+k\in \mathcal{P}\}$ where $\mathcal{P}$ is the set of primes, then $\sum _{s\in {\mathcal{S}}_k}\frac{1}{s}$ converges. This is what we will use in the proof.

As far as my knowledge goes (which is not too far), this proof is completely original, although that's not something to be very proud of!

Theorem : For any given $k$, there exists $k$ consecutive composite numbers.

Proof : If possible, let

$$\underset{p_1,p_2\text{ are consecutive primes}}{sup}|p_1-p_2|=B$$

for some integer $B$.

Then

$$\sum _{i=1}^B\sum _{p,p+i\in \mathcal{P}}\frac{1}{p}\ge \sum _{p\in \mathcal{P}}\frac{1}{p}$$

where $\mathcal{P}$ is the set of primes.

The LHS converges by Brun sieve and the RHS diverges, hence giving a contradiction! $◻$