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 $a_{n_0} \geq N$.

  • $\begingroup$ In your last statement, the $\forall n\in\mathbb N$ isn't meaningful - there isn't an $n$ in the rest of the statement. $\endgroup$ Apr 10, 2013 at 16:33
  • $\begingroup$ @GregMartin: typo fixed $\endgroup$
    – Lei Lei
    Apr 10, 2013 at 18:19

2 Answers 2


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 $k\ge k_0$.

  • $\begingroup$ probably the proof says that without loss of generality you can ignore the initial terms that don't respect the condition $\endgroup$
    – suissidle
    Apr 10, 2013 at 6:27
  • $\begingroup$ So $n_k$ from my example does not satisfies the hypothesis of the stated problem? $\endgroup$
    – Lei Lei
    Apr 10, 2013 at 7:23
  • $\begingroup$ Definitely not. $\endgroup$ Apr 10, 2013 at 16:34
  • $\begingroup$ @GregMartin: I guess I know why. I interpret $\lim a_k = \infty$ as $\{a_k\}_{k \geq 1}$ as a divergent sequence, not as a convergent sequence in extended real number system. $\endgroup$
    – Lei Lei
    Apr 10, 2013 at 18:18
  • $\begingroup$ Good diagnosis: the symbols $\lim a_k=\infty$ indicates a sequence that diverges in a very specific way. $\endgroup$ Apr 10, 2013 at 18:28

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

  • $\begingroup$ that is the definition for convergent sequence. I think for divergent sequence, it is $\forall n \exists n_0 > n$ such that $n_{k+1} > 3n_k n_{k-1} \ldots n_2 n_1$. $\endgroup$
    – Lei Lei
    Apr 10, 2013 at 7:39
  • $\begingroup$ you can consider infinity as a number .but in fact we say a same thing $\endgroup$
    – Xiaolang
    Apr 10, 2013 at 8:00
  • $\begingroup$ your definition fails for the example I constructed, but my definition still works. $\endgroup$
    – Lei Lei
    Apr 10, 2013 at 16:03

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