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 If $\frac {a_{(n+1)} }{a_n} < \frac {n^2}{(n+1)^2}$ and if $a_n > 0$ for all n . Then what can we say about the series $\sum a_n$?
Can anyone please help me by giving some hints?  I can not find any counter example to show that the series may not converge. Is it true?
 A: 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.
A: 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: 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.
