TITLE: 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
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: 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.