# Prove that $\sum{\frac{1}{n_k}}$ is convergent .

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?

• What if $m\leq 1$? Overall, the idea does work, but needs to be smoothed out a bit. May 26, 2020 at 17:04
• How do I show that $m \ne 1$ @MichaelBurr May 26, 2020 at 17:40
• 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...) May 26, 2020 at 17:48
• @DavidC.Ullrich why will $n_k$ be $1/2$ it is a positive integer. I didn't get you. May 26, 2020 at 17:51
• 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. May 26, 2020 at 18:02