A series of rational number converges to an irrational number

Question

This problem is an example from Putnam and Beyond, page 118:

Let$(n_k)_{k\geq1}$ be a strictly increasing sequence of positive integers with the property that

$$\displaystyle\lim_{k \to \infty} \frac{n_k}{n_1n_2 \ldots n_{k-1}} = \infty$$

Prove that the series $\sum_{k>1}\frac{1}{n_k}$ is convergent and the its sum is an irrational number.

The proof claims that $n_{k+1} \geq 3 n_k$ for $k\geq 3$ (Edit: for $k$ sufficiently large) and then ratio test to prove the convergence of $\sum_{k>1}\frac{1}{n_k}$.

My question: If I define $n_k$ as $n_1 = 1, n_2 = 2$ and $n_{2m+1} = m\times \prod_{i=1}^{2m} n_i, n_{2m+2}=n_{2m+1}+1$ for $m \geq1$ so that $\lim\sup \frac{n_k}{n_1n_2 \ldots n_{k-1}} = \infty$ and $\lim\inf \frac{n_k}{n_1n_2 \ldots n_{k-1}} = 0$, then does it satisfy the condition?

If we use $\{n_k\}$ constructed above, I can not see why $n_{k+1} \geq 3 n_k$ for $k$ sufficiently large.

Edit: My $\epsilon-\delta$ interpretation of $\lim a_k = \infty$ is that $\forall N \in \mathbb{R}^{+} \text{ and } \forall n \in \mathbb{N}, \exists n_0 \ge n $ such that $a_{n_0} \geq N$.

Answer 1

With your definition, it's clearly true that $\liminf \frac{n_k}{n_1n_2\cdots n_{k-1}} \le 1$. But that doesn't seem relevant to the stated problem, given its hypothesis.

It's not necessarily true that $n_{k+1} \ge 3n_k$ for $k\ge3$ (the sequence can start with $n_k=k$ for as many terms in a row as you like), but there does exist $k_0$ such that it's true for $k\ge k_0$.

Answer 2

let $\varepsilon=3$ then with the definition of limit there must be a $n_0\in N$ such that for any $n>n_0$ there has $n_{k+1}>3n_kn_{k-1}n_{k-2}...n_2n_1>3n_k$

and $n_0 = 3$ has no effects on how solving this problem