Does there exist a non-increasing sequence with these properties?

Question

Following on from Does there exist a sequence with these properties?

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

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

Note that the first obvious guesses - $a_n = 1/n$, $a_n = \log(n)/n$ don't work.

Probably one of those obnoxious "provide a context" objections, as if an interesting problem with an showing of work isn't enough. — marty cohen, Jan 28 '18 at 06:35
@Masacroso no, see the wolfram link. or google "nonincreasing function". or if its a hangup for you, just pretend i said "for which $n<m \implies a_m\leq a_n$". — user525678, Jan 28 '18 at 06:58
@Moti $\sum 1/n^2$ converges, so the partial sums are bounded. so the limit I wrote is $0$. — user525678, Jan 28 '18 at 07:01
The values go to zero but the sum is converging to a value greater than zero. May be you want to modify the condition. — Moti, Jan 28 '18 at 07:03
@Moti what is $\lim_{N\rightarrow \infty} \frac{1}{N}\sum_{n=1}^N 1/n^2$? — user525678, Jan 28 '18 at 07:12
@Moti he is right, the value of the first limit converges to zero when we choose $a_n:=1/n^2$ — , Jan 28 '18 at 07:15

score 5 · Answer 1 · answered Jan 30 '18 at 09:13

The answer is that there is no such sequence. If your sequence $a_n$ is non-increasing and bounded below (in this case by 0), it is a convergent sequence, so we have $a_n \to a$ for some $a \geq 0$. However, if a sequence converges, so does the sequence of it means (see for example this question), with the same limit. So we have

$$\liminf_{N\to\infty} \frac{1}{N} \sum_{n=1}^N a_n = \lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N a_n = a$$

Thus we need $a > 0$ to satisfy the first condition. But since $a_n$ is non-increasing, we also have $a_n \geq a$ for all $n\in \mathbb{N}$. Then $\frac{1}{a_n n^2} \leq \frac{1}{a n^2}$, and so we get

$$\sum_{n=1}^\infty \frac{1}{a_nn^2} \leq \sum_{n=1}^\infty \frac{1}{a n^2} = \frac{1}{a} \frac{\pi^2}{6} < \infty.$$

In other words, the second condition will not hold.

Does there exist a non-increasing sequence with these properties?

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