Contrast of 2 merging problems for series of non-negative numbers

Allow $a_n\geq 0$ be a series of non-negative numbers. Take into consideration the adhering to 2 declarations: $$ \text{(I)}\qquad\qquad \lim_{n\to\infty} \frac{1}{n^2}\sum_{i=1}^n a_i =0, $$. $$ \text{(II)}\qquad\qquad\qquad \sum_{n=1}^\infty \frac{a_n}{n^2}<\infty. $$

Inquiries: Does (I) indicate (II)? Does (II) indicate (I)? Or else, please give counterexamples.

Inspiration: Both declarations take place in the context of the law of large numbers for non-identically dispersed arbitrary variables. With $a_n=\mathrm{Var}(X_n)$, one can end the weak LLN if the $X_n$ are pairwise uncorrelated and also problem (I) holds. The solid LLN can be ended if the $X_n$ are stochastically independent and also problem (II) holds. Consequently, one could anticipate that (II) indicates (I).

0
2019-05-06 23:24:38
Source Share
Answers: 1

II indicates I :

(we manage the partial amounts below)

One uses the Cauchy - Schwarz inequality to get :

$\displaystyle \left( \sum_{i=1}^n \frac{a_i}{i^2} \right)\left(\sum_{i=1}^n i^2\right) \ge (\sum_{i=1}^n a_i)^2 $

We can see that $ \sum_{i=1}^n i^2 $ is of the size $n^3$ as $n \rightarrow \infty$. So when we separate both sides by $n^4$, LHS will certainly merge to 0 as $n \rightarrow 0$, which indicates that $\displaystyle \frac{\sum a_n}{n^2}$ merges to 0.

I does NOT indicate II :

as an example take $a_1,a_2$ to be anything, $\displaystyle a_n = \frac{n}{\log n}$ for $n \ge 3$. This does not please II (popular). Yet it does please I, due to the fact that asymptotically, $\displaystyle \frac{1}{n^2} \sum_{i=1}^n a_i \sim \frac{1}{n^2} \int_3^n \frac{x}{\log x} dx$.

The logarithmic integral has actually a popular estimate $\displaystyle \int_3^n \frac{1}{\log x}dx = O\left(\frac{n}{\log n}\right)$.

So $\displaystyle \frac{1}{n^2} \int_3^n \frac{x}{\log x} dx \leq \frac{n}{n^2} \int_3^n \frac{1}{\log x} dx = O\left(\frac{1}{\log n}\right) \rightarrow 0$ as $n \rightarrow \infty$.

0
2019-05-08 19:26:49
Source