<s>[INST] conditions on $\{a_n\}$ that imply convergence of $\sum_{n=1}^{\infty} a_n$ (NBHM 2011) Question is :
For a sequence $\{a_n\}$  of positive terms, Pick out the cases which imply convergence  of $\sum_{n=1}^{\infty} a_n$.


*

*$\lim_{n\rightarrow \infty} n^{\frac{3}{2}}a_n=\frac{3}{2}$

*$\sum_{n=1}^{\infty} n^2a_n^2<\infty$

*$\dfrac{a_{n+1}}{a_n}< (\frac{n}{n+1})^2$


For the first case $\lim_{n\rightarrow \infty} n^{\frac{3}{2}}a_n=\frac{3}{2}$ :
Suppose $a_n > \frac{3}{2}. (\frac{1}{n})^{\frac{3}{2}}$ for all $n\in \mathbb{N}$ then we would have :
$n^{\frac{3}{2}}a_n > \frac{3}{2}.n^{\frac{3}{2}} (\frac{1}{n})^{\frac{3}{2}}=\frac{3}{2}$ but then $n^{\frac{3}{2}}a_n$ would not converge to $\frac{3}{2}$
So, for large $n$ we should have $$a_n \leq\frac{3}{2}. \frac{1}{n^{\frac{3}{2}}}\Rightarrow \sum_{n=1}^{\infty}a_n \leq \sum_{n=1}^{\infty} \frac{3}{2}. \frac{1}{n^{\frac{3}{2}}}=\frac{3}{2}\sum_{n=1}^{\infty}  \frac{1}{n^{\frac{3}{2}}}$$ 
Right hand side converges and by comparison test $\sum_{n=1}^{\infty}a_n$ should converge.
For the second case $\sum_{n=1}^{\infty} n^2a_n^2<\infty$ :
Suppose that $na_n> \dfrac{1}{\sqrt{n}}$ for all $n$ then we would have $$n^2a_n^2> \dfrac{1}{n}$$ but then it is given that $\sum_{n=1}^{\infty} n^2a_n^2$ which would imply that $\sum_{n=1}^{\infty}\dfrac{1}{n}$ converges which is a contradiction.
Thus we should have $na_n\leq \dfrac{1}{\sqrt{n}}$ i.e., $a_n\leq \frac{1}{n\sqrt{n}}=\frac{1}{n^{\frac{3}{2}}}$ for all $n$ large.
So, We have $$\sum_{n=1}^{\infty}a_n\leq\sum_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}$$
right hand side is convergent so by comparison test $\sum_{n=1}^{\infty}a_n$ converges.
For third case $\dfrac{a_{n+1}}{a_n}< (\frac{n}{n+1})^2$ :
we would have $\lim_{n\rightarrow \infty} \dfrac{a_{n+1}}{a_n} <\lim_{n\rightarrow \infty} (\frac{n}{n+1})^2=1$
Thus, by ration test, $\sum_{n=1}^{\infty} a_n$ Converges absolutely and so is convergent.
My choice of bounds :


*

*$a_n > \frac{3}{2}. (\frac{1}{n})^{\frac{3}{2}}$ for all $n$ in first case

*$na_n> \dfrac{1}{\sqrt{n}}$ for all $n$ in second case


All these came to me just by choice and i would like to know if there is some sense behind this choice.
Please help me to clear this and make this solution a bit more clear.
Please do not give an alternative solution
 until this problem is fully verified.
Thank you.
 [/INST] Comment on your first case:
$n^{1.5}a_n$ can converge to 1.5 while always staying above 1.5. But it must eventually go below (say) 3 so that eventually $a_n \leq 3/n^{1.5}$.
Comment on your third case:
You wrote since $a_n \geq 1/n^2$ (which isn't true anyway), comparison test applies which is also false. What you really get is $a_n \leq a_1/n^2$ (easily verified) which is enough to give convergence.
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