If $\frac {a_{(n+1)} }{a_n} < \frac {n^2}{(n+1)^2}$ and if $a_n > 0$ for all n . $\sum a_n$ converges or not

Write the given inequality as

$$ a_{n+1} < (n^2 a_n)/(n+1)^2.$$

Then all it takes is to iterate the above, i.e., using that

$$ a_n < ((n-1)^2a_{n-1})/n^2,$$

We arrive at

$$ a_{n+1} < ((n-1)^2 a_{n-1})/(n+1)^2.$$

By now it should be clear that we can continue up until we reach the first term of the sequence. In other words, we may prove by induction that

$$ a_{n+1} < a_1/(n+1)^2.$$

Of course, the series formed by the terms on the RHS in the above inequality converges, and therefore, by comparison, so does the original one.

More generally, suppose $\dfrac {a_{n+1} }{a_n} \lt \dfrac {n^c}{(n+1)^c} $ where $c > 0$.

Then

$\begin{array}\\ \dfrac{a_n}{a_1} &=\prod_{k=1}^{n-1}\dfrac {a_{k+1} }{a_k}\\ &<\prod_{k=1}^{n-1}\dfrac{k^c}{(k+1)^c}\\ &=\dfrac{\prod_{k=1}^{n-1}k^c}{\prod_{k=1}^{n-1}(k+1)^c}\\ &=\dfrac{\prod_{k=1}^{n-1}k^c}{\prod_{k=2}^{n}k^c}\\ &=\dfrac{1^c\prod_{k=2}^{n-1}k^c}{n^c\prod_{k=2}^{n-1}k^c}\\ &=\dfrac{1}{n^c}\\ \end{array} $

so that $\sum_{n=1}^m a_n \lt a_1\sum_{n-1}^m \dfrac1{n^c} $ and this converges when $c > 1$ (integral test, zeta function, etc.).

So it certainly converges when $c = 2$.

A bruteforce approach: since $\displaystyle \log a_{n+1}-\log a_n < \log\left(\frac{n^2}{(n+1)^2} \right)$ and $\displaystyle \log\left(\frac{n^2}{(n+1)^2} \right) = -\frac 2n +O\left(\frac 1{n^2} \right)$ there exists some $M>0$ such that $\displaystyle \forall n\geq 1, \log\left(\frac{n^2}{(n+1)^2} \right)\leq -\frac 2n +\frac M{n^2}$

By summing there exists some $M'$ and $M''$ such that $$\log a_{n+1}-\log a_1 < -2H_n+M'\leq -2\log n +M''$$

Thus there is some $M'''$ such that $$\begin{align}\log a_{n+1}&\leq -2\log n +M'''\\ \implies a_{n+1}&\leq \exp(-2\log n)\exp(M''') \\ \implies a_{n+1}&\leq \frac 1{n^2}\exp(M''') \end{align}$$

Hence $\sum a_n$ converges.