Does there exist a sequence with these properties?

Question

Does there exist a sequence $a_n\in \mathbb{R}^{>0}$ such that $$\liminf_{N\rightarrow \infty}~~~ \frac{1}{N}\sum_{n=1}^Na_n >0$$

and $$\sum_{n=1}^\infty \frac{1}{a_n\cdot n^2} = \infty~~?$$

The second condition is impossible for $a_n\in \Bbb N$ because then $0<1/(a_nn^2)\leq 1/n^2 $ and $\sum_{n=1}^{\infty}(1/n^2)<\infty.$ — DanielWainfleet, Jan 28 '18 at 05:57

score 2 · Accepted Answer · answered Jan 28 '18 at 05:58

2

Let $a_n= 1/n^2$ when $n$ is even, and $1$ when $n$ is odd. Then the average converges to $1/2$ and the sum is $\infty$.

answered Jan 28 '18 at 05:58

saulspatz

The OP asks for a sequence of natural numbers, though. – G Tony Jacobs Jan 28 '18 at 06:00
@GTonyJacobs Oops. Gotta read all the words. – saulspatz Jan 28 '18 at 06:02
1

@GTonyJacobs sorry. my mistake. the second condition doesn't make sense for natural numbers. – user525678 Jan 28 '18 at 06:02
What I see in the Q is $\sum (a_n n^2)^{-1}.$ For $a_n\in \Bbb N$ we have $0<(a_nn^2)^{-1}\leq n^{-2}$ so the sum must be finite because $\sum_{n=1}^{\infty}n^{-2}=\pi^2 /6$. – DanielWainfleet Jan 28 '18 at 06:04
@DanielWainfleet When $n$ is even, $a_nn^2 = 1,$ so the summand is $1$ infintely often. – saulspatz Jan 28 '18 at 06:07
@saulspatz thanks for your answer. I realized that I mean to ask about non-increasing sequences, but this is illuminating nonetheless (i've now asked a seperate question in a new post). – user525678 Jan 28 '18 at 06:12
The Q has been re-defined by an edit. My prior comments were toward the first version, which said $a_n\in \Bbb N$. – DanielWainfleet Jan 28 '18 at 06:35
@DanielWainfleet Ah, now your comments make sense to me. I was rather perplexed. – saulspatz Jan 28 '18 at 06:48

1 Answers1