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The sequence $(n_k)$ is a strictly increasing positive sequence of integers which satisfies the condition

$\lim_{k\to\infty}\frac{n_k}{n_1...n_{k-1}}=\infty$.

Now this has been my attempt,

$\frac{n_k}{n_1...n_{k-1}}>1$ for all $k \ge N_1$. So $n_k > n_1...n_{k-1}$ for all $k \ge N_1$.

Since the sequence is strictly increasing so $(n_k) >M_{N_1}$ for all $k \ge N_1$ ,then $\frac{1}{n_k}<\frac{1}{M_{N_1}}$ for all $k \ge N_1$.Now let us chose $\min m=(n_1,...,n_{N_1-1},{M_{N_1}})$.

Then $n_k> m$ for all $k \in N$

Now $\sum{\frac{1}{n_k}}<\sum{\frac{1}{n_1...n_{k-1}}}<\sum{\frac{1}{m^{k-1}}}$ for $ k \ge N_1$ which leads to a geometric sequence.

I think this method should work.It would be very helpful if someone goes through my attempt and point out my mistake.

Edit1:The question also has a subpart which ask to prove that the sum is irrational.How do I proceed?

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    $\begingroup$ What if $m\leq 1$? Overall, the idea does work, but needs to be smoothed out a bit. $\endgroup$ May 26, 2020 at 17:04
  • $\begingroup$ How do I show that $m \ne 1$ @MichaelBurr $\endgroup$
    – Antimony
    May 26, 2020 at 17:40
  • $\begingroup$ This is false as stated; $n_k=1/2$ is a counterexample. Probably you were supposed to assume that $n_k$ is an integer? (In any case, your solution must be wrong since it's supposedly a proof of something false...) $\endgroup$ May 26, 2020 at 17:48
  • $\begingroup$ @DavidC.Ullrich why will $n_k$ be $1/2$ it is a positive integer. I didn't get you. $\endgroup$
    – Antimony
    May 26, 2020 at 17:51
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    $\begingroup$ Then case closed! your attempt seems to work fine. In fact, you may assume, without loose of generality that $n_1\geq2$ so that $n_k>2^{k-1}$ for all $k$ large enough. $\endgroup$
    – Mittens
    May 26, 2020 at 18:02

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