Proving that the limit is bounded. 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!
 A: 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).
