I am trying the following exercise that's actually stated as a lemma on an Econometrics textbook:
Let $\{a_t\}_{t = 1,2.,..}$ be a sequence of nonnegative scalars such that $\sum\limits_{t = 1}^{T}\frac{a_t}{T} < M < \infty$ for every $T$ for some finite $M$. Prove that $\lim\limits_{T\rightarrow \infty}\sum\limits_{t = 1}^{T}\frac{a_t}{t^2} < \infty$.
I tried using the Cauchy-Schwarz inequality but I didn't take me very far. So I did this trick, although I'm not totally sure I'm taking a safe path:
For any given $T$:
$\sum\limits_{t = 1}^{T}\frac{a_t}{t^2} = \sum\limits_{t \leq \sqrt{T}}\frac{a_t}{t^2} + \sum\limits_{t > \sqrt{T}}\frac{a_t}{t^2} \leq \sum\limits_{t \leq \sqrt{T}} a_t + \sum\limits_{t > \sqrt{T}}\frac{a_t}{T}$
The second term cannot be bigger than $M - M\sqrt{T} + 1$. The first term is smaller than $M\sqrt{T}$. Then this sum is bounded for any $T$ and we would have the desired result.
I would appreciate some comments on if this is true or if I am missing something! Thanks a lot in advance!
We have that $$ \forall T \in \mathbb{N}^+ \qquad \sum_{i=1}^{T}a_i < MT. $$ We can define $s_i$ as the cumulative sum, i.e., $$ s_i = \sum_{j=1}^{i} a_j $$ and we notice that $$ A \left( \begin{array}{c} s_1\\ s_2\\ \vdots\\ s_T \end{array} \right ) = \left( \begin{array}{c} a_1\\ a_2\\ \vdots\\ a_T \end{array} \right ). $$ Where $$ A=\left ( \begin{array}{cccc} 1 & 0 & & 0\\ -1 & \ddots & \ddots & \\ & \ddots & \ddots & 0 \\ 0 & & -1 & 1 \end{array} \right )$$ Therefore $$ \sum_{i=1}^{T}\frac{a_i}{i^2} = \left( \begin{array}{cccc} 1 & \frac{1}{2^2} & \cdots & \frac{1}{T^2} \end{array} \right) \left( \begin{array}{c} a_1\\ a_2\\ \vdots\\ a_T \end{array} \right ) = \left( \begin{array}{cccc} 1 & \frac{1}{2^2} & \cdots & \frac{1}{T^2} \end{array} \right) A \left( \begin{array}{c} s_1\\ s_2\\ \vdots\\ s_T \end{array} \right ). $$ By associativity we rewrite the expression as $$ \sum_{i=1}^{T}\frac{a_i}{i^2} = \left( \begin{array}{cccc} 1 - \frac{1}{2^2} & \frac{1}{2^2} -\frac{1}{3^2} & \cdots & \frac{1}{T^2} \end{array} \right) \left( \begin{array}{c} s_1\\ s_2\\ \vdots\\ s_T \end{array} \right ) < \left( \begin{array}{cccc} 1 - \frac{1}{2^2} & \frac{1}{2^2} -\frac{1}{3^2} & \cdots & \frac{1}{T^2} \end{array} \right) \left( \begin{array}{c} M\\ 2 \cdot M\\ \vdots\\ T \cdot M \end{array} \right ) . $$ Where the inequality follows from the fact that all the components of the horizontal vector are positive.
The RHS can be rewritten again as $$ \left( \begin{array}{cccc} 1 & \frac{1}{2^2} & \cdots & \frac{1}{T^2} \end{array} \right) A \left( \begin{array}{c} M\\ 2 \cdot M\\ \vdots\\ T \cdot M \end{array} \right ) = M \cdot \left( \begin{array}{cccc} 1 & \frac{1}{2^2} & \cdots & \frac{1}{T^2} \end{array} \right) \left( \begin{array}{c} 1\\ 1\\ \vdots\\ 1 \end{array} \right ) = M \sum_{i=1}^{T}\frac{1}{i^2}<M\frac{\pi^2}{6} $$ where the last inequality is a well known result (https://en.wikipedia.org/wiki/Basel_problem).