Many people would know that the series $$\sum_{n=1}^{\infty}\frac{1}{p_n}$$ where $p_n$ is the $n^{\text{th}}$ prime number, diverges. While reading Tom Apostol's Introduction to analytic number theory, I found a short proof (due to Clarkson) that the series diverges. Here it is:
We assume the series converges and obtain a contradiction. If the series converges there is an integer $k$ such that $$\sum_{m=k+1}^{\infty} \frac{1}{p_m}<\frac{1}{2}$$ Let $Q = p_1...p_k$ , and consider the numbers $1 + nQ$ for $n = 1, 2, . ..$ None of these is divisible by any of the primes $p_1 , ... , p_k$ . Therefore, all the prime factors of $l ± nQ$ occur among the primes $p_{k+1}, p_{k+ 2}, . . .$ Therefore for each $r \geq 1$ we have $$\sum_{n=1}^{r}\frac{1}{1+nQ}\leq\sum_{t=1}^{\infty}\left(\sum_{m=k+1}^{\infty}\frac{1}{p_m}\right)^t$$ since the sum on the right includes among its terms all the terms on the left. But the right-hand side of this inequality is dominated by the convergent geometric series $$\sum_{t=1}^{\infty}\left(\frac{1}{2}\right)^t$$ Therefore the series $\sum\limits_{n=1}^{\infty}1/(1+nQ)$ has bounded partial sums and hence converges. But this is a contradiction because the integral test or the limit comparison test shows that this series diverges.
As I mentioned in my infinitude of primes question, I mentioned that I am not content with only one proof. So please give more proofs of the divergence of the sum of reciprocals. I am not the type of person who thinks that long proof are not elegant; whether short or long, I like every proof. If the proof is too long for this site, please link an article containing the proof.