# Bounded Mean of Partial Sums and Convergence

This is from Amemiya's Advanced Econometrics Chapter 3 Q5:

Suppose $\{a_t\}$ is a non-negative sequence such that $\frac{1}{T}\sum_{t=1}^{T} a_t < M$ for all $T$. Prove that $\lim_{T \rightarrow \infty} \sum_{t=1}^{T} \frac{a_t}{t^2} < \infty$.

I'm having problems using the supposition. I have tried to break up either series as such:

$\frac{1}{T^2} \sum_{t=1}^{T^2} a_t \geq \frac{1}{T^2} \sum_{t=1}^{T} a_t + \frac{1}{T^2} \sum_{t=T+1}^{T^2} a_t \geq \frac{1}{T^2} \sum_{t=1}^{T} a_t + \sum_{t=T+1}^{T^2} \frac{a_t}{t^2}$

However, the bounds are not tight enough. I have also tried to redefine the supposition so as to apply Kronecker's lemma, but the results are usually a mess.

Would appreciate any help on the matter. My ultimate goal is to apply Kolmogorov's SLLN and chanced upon this result that I would like to use.

## 1 Answer

Let $A_T = \sum_{t=1}^T a_t$.

Summation by parts yields

$$\sum_{t=1}^T \frac{a_t}{t^2} = \frac{A_T}{(T+1)^2} + \sum_{t=1}^TA_t\left( \frac{1}{t^2} - \frac{1}{(t+1)^2}\right) \\ = \frac{A_T}{T}\frac{T}{(T+1)^2} + \sum_{t=1}^T\frac{A_t}{t}\left( \frac{2t+1}{t(t+1)^2} \right).$$

The first term on the RHS converges to $0$ as $T \to \infty$ since $A_T/T < M$. Furthermore, the second term is a convergent series by the comparison test, since

$$\frac{A_t}{t}\left( \frac{2t+1}{t(t+1)^2}\right) \leqslant \frac{3M}{(t+1)^2},$$

and $\sum_{t=1}^\infty (t+1)^{-2}$ converges.