Does $\sum \frac{a_n}{1+n a_n}$ converge if $a_n = \frac{1}{\sqrt{n}}$ if $n$ is a perfect square and $a_n = \frac{1}{n^2}$ otherwise? Given the sequence $\left\{ a_n \right\}$, where 
$$a_n = \begin{cases} \frac{1}{\sqrt{n}} \ \mbox{ if } n \mbox{ is a perfect square} \\ \frac{1}{n^2} \ \mbox{ otherwise}, \end{cases}$$ 
does the series $$\sum \frac{a_n}{1 + n a_n}$$ converge or diverge? 
My effort: 
If $n = m^2$, then we have 
$$\frac{a_n}{1+na_n} = \frac{ \frac{1}{m} }{ 1 + m^2 \frac{1}{m} } = \frac{1}{m(m+1)} = \frac{1}{\sqrt{n} (1 + \sqrt{n})}, $$ 
and otherwise 
$$ \frac{a_n}{1+na_n} = \frac{ \frac{1}{n^2}}{1+ n \frac{1}{n^2}} = \frac{1}{n^2 + n}. $$ 
Thus we have 
$$ \frac{a_n}{1+na_n} = \begin{cases} \frac{1}{\sqrt{n} (1 + \sqrt{n})} \ \mbox{ if } n \mbox{ is a perfect square} \\ \frac{1}{n( 1 + n)} \ \mbox{ otherwise}. \end{cases} $$
What next? How to proceed from here? 
An afterthought:
If $n$ is a perfect square, then we note that 
$$\frac{a_n}{1+na_n} = \frac{1}{\sqrt{n}(1+ \sqrt{n})} \geq \frac{1}{2n},$$

Hint.
$$\sum_{n=1}^\infty\frac{a_n}{1+na_n}=\sum_{k=1}^\infty\frac{a_{k^2}}{1+k^2 a_{k^2}}+\sum_{n\text{ is not a square}} \frac{a_n}{1+na_n}$$
From what you've done we know
$$\sum_{k=1}^\infty\frac{a_{k^2}}{1+k^2 a_{k^2}}=\sum_{k=1}^\infty\frac 1{k(k+1)}<\infty$$
and
$$\sum_{n\text{ is not a square}} \frac{a_n}{1+na_n}=\sum_{n\text{ is not a square}}\frac 1{n(n+1)}<\sum_{n=1}^\infty\frac 1{n(n+1)}<\infty$$
