# ${p_n}$ is such that $S_n = {1\over p_1}+{1\over p_2}+\cdots+{1\over p_n}$ converges. Prove $\sigma_n = \left(1+{1\over p_1}\right)\cdots$ converges

Let $$\{p_n\}$$ denote a sequence such that: $$S_n = {1\over p_1} + {1\over p_2} + \cdots + {1\over p_n}$$ converges. Prove that: $$\sigma_n = \left(1+{1\over p_1}\right)\left(1+{1\over p_2}\right)\cdots\left(1+{1\over p_n}\right)$$ converges, where $$n, p_n \in \Bbb N$$.

Consider each bracket from $$\sigma_n$$. By $${1\over p_n} > 0$$: $$\forall k \in \Bbb N: \left(1+{1\over p_k}\right) > 1$$ So $$\sigma_n$$ must be monotonically increasing by: $${\sigma_{n+1} \over \sigma_n} = \left(1+{1\over p_{n+1}}\right) > 1$$

To show a monotonic sequence is convergent it's sufficient to show that it's bounded above. Lets try to find the bound. Recall: $$\ln(1+x) \le x \iff (1+x) \le e^x \tag1$$ So by $$(1)$$ we have: $$\sigma_n \le e^{S_n}$$ But $$S_n$$ is convergent! And thus: $$\sigma_n \le e^L$$ where: $$L = \lim_{n\to\infty}S_n$$ Which by monotone convergence theorem proves $$\sigma_n$$ is convergent.

I'm kindly asking to verify my proof and point to the mistakes in case of any. Thank you!

• If $p_n >0$, $\sigma_n$ is easily proved to be increasing and the inequality $\sigma_n \leq e^{S_n} \leq e^L$ is easy as well. Some passages are very strange (why do you need $1/p_n \rightarrow 0$? What is your digression about the harmonic series?) but the crux seems quite correct. – Mindlack Jan 18 at 14:06
• @Mindlack please notice a part in the question section saying $n, p_n \in \Bbb N$ – roman Jan 18 at 14:09
• @Mindlack as far as the other notices, I agree with you, I will update the post – roman Jan 18 at 14:10
• Right, thank you. I edited. – Mindlack Jan 18 at 14:11
• This is correct now. You can actually even write that each factor $(1+1/p_n)$ is lower than $e^{1/p_n}$, shortening the argument. – Mindlack Jan 18 at 14:16