Convergence of the series with $a_n= \frac{1}{\sqrt{1} \cdot n^2} + \frac{1}{\sqrt{2} \cdot (n-1)^2} +\cdot\cdot\cdot + \frac{1}{\sqrt{n} \cdot 1^2}$ I need to determine the convergence of the series whose general term is given by:
$$a_n= \frac{1}{\sqrt{1} \cdot n^2} + \frac{1}{\sqrt{2} \cdot (n-1)^2} +\cdot\cdot\cdot + \frac{1}{\sqrt{n} \cdot 1^2}$$
Observation

*

*$\frac{1}{n^\frac{3}{2}} \lt a_n \lt \sum_{k=1}^n \frac{1}{k^2 }$

*$a_n$ is monotonic decreasing sequence and converges to zero.

This series is bigger than a convergent and smaller than a divergent series by the second observation, if I have not incorrectly done. But its of no help here to determine the convergence. Kindly help, thanks in advance.
 A: Let $N \in \mathbb{N}$. One has
\begin{align*}
\left( \sum_{k=1}^N \frac{1}{\sqrt{k}}\right)\left( \sum_{j=1}^N \frac{1}{j^2}\right) &= \sum_{k=1}^N \sum_{j=1}^N \frac{1}{j^2\sqrt{k}}\\
&\leq \sum_{n=2}^{2N} \sum_{j+k=n} \frac{1}{j^2\sqrt{k}}\\
&= \sum_{n=1}^{2N-1} \sum_{j+k=n+1} \frac{1}{j^2\sqrt{k}}\\
&= \sum_{n=1}^{2N-1} \sum_{k=1}^{n} \frac{1}{\sqrt{k}(n+1-k)^2}\\
\end{align*}
So you get that
$$\sum_{n=1}^{2N-1} a_n \geq \left( \sum_{k=1}^N \frac{1}{\sqrt{k}}\right)\left( \sum_{j=1}^N \frac{1}{j^2}\right) $$
and because $\displaystyle{\sum \frac{1}{\sqrt{k}}}$ is divergent and $\displaystyle{\sum \frac{1}{j^2}}$ is convergent, you get that
$$\boxed{\text{the series }\sum a_n \text{ diverges}.}$$
Edit : This gives a precise estimation of the partial sum of the series $\sum a_n$, but as @PeterSzilas noticed, the divergence can be obtained much more directly by noticing that
$$a_n \geq \frac{1}{\sqrt{n}}$$
A: $S_N= \sum_{n=1}^N \sum_{k=1}^n \frac{1}{\sqrt{k}(n+1-k)^2}$
we have
$1\leq k \leq n\leq N$
So
$S_N=\sum_{k=1}^N \sum_{n=k}^N \frac{1}{\sqrt{k}}\frac{1}{(n+1-k)^2}$
$= \sum_{k=1}^N\frac{1}{\sqrt{k}} \left(\sum_{n=k}^N \frac{1}{(n+1-k)^2}\right)$
But
$\sum_{n=k}^N \frac{1}{(n+1-k)^2}\geq \int_{1}^{N-k+2} \frac{1}{x^2}dx
\geq 1-\frac{1}{N-k+2}\geq\frac12$
$S_N\geq \frac{1}{2}\sum_{k=1}^N\frac{1}{\sqrt{k}}$
