Example of series such that $\sum a_n$ is divergent but $\sum \frac{ a_n}{1+ n a_n}$ is convergent 
Example of series such that $\sum a_n$ is divergent but $\sum \frac{ a_n}{1+ n a_n}$ is convergent.

I got one example from online that $a_n =\frac {1}{n^2} $for n is non square term and  $a_n =\frac {1}{\sqrt n} $ for $n$ is square term . I know that $\sum a_n$ is divergent term but not convience with  convergence of  $ \sum \frac{ a_n}{1+ na_n}$. Any help will be appreciated.
 A: The counterxample is correct. If $a_n =\frac {1}{n^2}$ when $n$ is non square term and  $a_n =\frac {1}{\sqrt n} $ when n is square term, we have that
$$\sum_{n=1}^{\infty}  a_n=\sum_{\text{$n$ is not a square}} \frac{1}{n^2}+\sum_{\text{$n$ is a square}} \frac{1}{\sqrt{n}}\geq  \sum_{k=1}^{\infty} \frac{1}{\sqrt{k^2}}=+\infty.$$
One the  other hand 
$$\sum_{n=1}^{\infty} \frac{ a_n}{1+ n a_n}=\sum_{\text{$n$ is not a square}} \frac{1}{n^2+ n}+\sum_{\text{$n$ is a square}} \frac{1}{\sqrt{n}+ n}\\\leq \sum_{n=1}^{\infty} \frac{1}{n^2}+\sum_{k=1}^{\infty} \frac{1}{k+ k^2}\leq 2\sum_{n=1}^{\infty} \frac{1}{n^2}<+\infty$$
A: Delta-u and Robert Z have shown how to show convergence of $\sum{a_n\over1+na_n}$ for the example the OP was considering. Another example along the same lines is
$$a_n=\cases{1\quad\text{if }n=2^m\text{ for some } m\in\mathbb{N}\\
0\quad\text{otherwise}}$$
In this case the sum $\sum a_n$ diverges because $a_n$ does not converge to $0$ (because it's infinitely often equal to $1$), while
$$\sum_n{a_n\over1+na_n}=\sum_m{1\over1+2^m}\lt\sum_m{1\over2^m}=2$$
The key idea in all the examples is to take a divergent series (like $1+{1\over2}+{1\over3}+\cdots$, in the OP's example, or, as here, $1+1+1+\cdots$) and space its terms more and more widely apart, with convergently small stuff (e.g., $1+{1\over4}+{1\over9}+\cdots$ or just plain $0+0+0+\cdots$) in between.
A: Hint
To show that $\sum \frac{ a_n}{1+ n \cdot a_n}$ is convergent notice that (as the terms are positive such decomposition is allowed):
 $$\sum \frac{ a_n}{1+ n \cdot a_n}=\sum_{n \text{ non-square}}  \frac{\frac{1}{n^2}}{\frac{1}{n^2}+n\frac{1}{n^2}}+\sum_{m} \frac{\frac{1}{m}}{1+m^2\frac{1}{m}}=\sum_{n \text{ non-square}}  \frac{1}{n^2+n}+\sum_{m} \frac{1}{m+m^2}$$
A: Slightly easier solution: Let $S=\{m^2:m\in \mathbb N\}.$ Define $a_n = 1$ if $n\in S,$ $a_n = 0$ otherwise. Then $\sum a_n$ diverges, since $a_n =1$ for infinitely many $n,$ hence $a_n\not \to 0.$ But
$$\sum_{n=1}^{\infty}\frac{a_n}{1+na_n} = \sum_{m=1}^{\infty} \frac{1}{1+m^2\cdot 1} \le \sum_{m=1}^{\infty} \frac{1}{m^2}<\infty.$$
