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!

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    $\begingroup$ 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. $\endgroup$ – Mindlack Jan 18 at 14:06
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    $\begingroup$ @Mindlack please notice a part in the question section saying $n, p_n \in \Bbb N$ $\endgroup$ – roman Jan 18 at 14:09
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    $\begingroup$ @Mindlack as far as the other notices, I agree with you, I will update the post $\endgroup$ – roman Jan 18 at 14:10
  • $\begingroup$ Right, thank you. I edited. $\endgroup$ – Mindlack Jan 18 at 14:11
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    $\begingroup$ 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. $\endgroup$ – Mindlack Jan 18 at 14:16

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