The following text is from the book Introduction to Analytic Number Theory by T. M. Apostol :
Theorem 1.13 $ \ $ The infinite series $\sum_{n=1}^\infty 1/p_n$ diverges.
Proof. The following short proof of this theorem is due to Clarkson [11]. 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 $Q=p_1\cdot\cdot\cdot p_k$, 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 $1+nQ$ occur among the primes $p_{k+1},p_{k+2},…$ Therefore for each $r\geq 1$ we have $$\sum_{n=1}^r\frac{1}{1+nQ}\leq\sum_{t=1}^\infty\left(\displaystyle\sum_{m=k+1}^\infty\frac{1}{p_m}\right)^t,$$ 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 $$\sum_{t=1}^\infty\left(\displaystyle\frac{1}{2}\right)^t.$$ Therefore the series $\sum_{n=1}^\infty 1/(1+nQ)$ 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.
Clear simple explanation about the following questions all from the last paragraph would be much appreciated :
$1-$ The sum $\sum_{n=1}^{r} (1+nQ)^{-1}$ is bounded and it was considered $r$ to be finite. How it implies a decision about $\sum_{n=1}^{\infty} (1+nQ)^{-1}$ i.e. $r \rightarrow \infty$? I don't understand one to the last sentence of the proof why if an infinite series has bounded partial sum then it converges; harmonic series has bounded partial sum but it doesn't converge.
$2-$ I know a bit of real analysis regarding the last sentence but what are the integrals and compared thing to test? How is the integral test in here? What is the limit comparison test in this example? And how they result to the divergence of $\sum_{n=1}^{\infty} (1+nQ)^{-1}$?
Best Answer
Referring, as you say, to the last three lines in the text:
The series diverges (which, of course, is a positive one) since for example
$$\frac1{1+nQ}\xrightarrow[n\to\infty]{}0\;\;\text{monotonically descending and}\;\;\int_1^\infty\frac{dx}{1+xQ}=\left.\frac1Q\log(1+xQ)\right|_1^\infty$$
clearly diverges, so does the series.
Another way to see it diverges:
$$\frac1{1+nQ}\ge\frac1{nQ+nQ}=\frac1{2Q}\frac1n$$
and the right hand is just a scalar multiple of the harmonic series and thus divergent, so by the comparison test our series diverges.