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.

Solution 1:

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 $$a_n \geq \frac{1}{\sqrt{n}}$$