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Let $a_n\geq 0$ be a sequence of non-negative numbers. Consider the following two statements: $$ \text{(I)}\qquad\qquad \lim_{n\to\infty} \frac{1}{n^2}\sum_{i=1}^n a_i =0, $$ $$ \text{(II)}\qquad\qquad\qquad \sum_{n=1}^\infty \frac{a_n}{n^2}<\infty. $$

Questions: Does (I) imply (II)? Does (II) imply (I)? Otherwise, please provide counterexamples.

Motivation: Both statements occur in the context of the law of large numbers for non-identically distributed random variables. With $a_n=\mathrm{Var}(X_n)$, one can conclude the weak LLN if the $X_n$ are pairwise uncorrelated and condition (I) holds. The strong LLN can be concluded if the $X_n$ are stochastically independent and condition (II) holds. Therefore, one might expect that (II) implies (I).

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1 Answer 1

up vote 11 down vote accepted

II implies I:

(we deal with the partial sums here)

One applies 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 magnitude $n^3$ as $n \rightarrow \infty$. So when we divide both sides by $n^4$, LHS will converge to 0 as $n \rightarrow 0$, which implies that $\displaystyle \frac{\sum a_n}{n^2}$ converges to 0.

I does NOT imply II:

e.g. take $a_1,a_2$ to be anything, $\displaystyle a_n = \frac{n}{\log n}$ for $n \ge 3$. This does not satisfy II (well known). But it does satisfy I, because 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 a well known approximation $\displaystyle \int_3^n \frac{1}{\log x}dx = O\left(\frac{n}{\log n}\right)$.

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 $n \rightarrow \infty$.

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Nice! Thank you for your help. –  Rasmus Aug 8 '10 at 13:03

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